Continuity Questions in English

(A) For $f(x)$ to be continuous at $x = \frac{\pi}{4}$,the left-hand limit must equal the right-hand limit:
$\lim_{x \rightarrow \frac{\pi}{4}^-} (x + a \sqrt{2} \sin x) = \frac{\pi}{4} + a \sqrt{2} \left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4} + a$
$\lim_{x \rightarrow \frac{\pi}{4}^+} (2x \cot x + b) = 2\left(\frac{\pi}{4}\right) \cot\left(\frac{\pi}{4}\right) + b = \frac{\pi}{2} + b$
Equating them: $\frac{\pi}{4} + a = \frac{\pi}{2} + b \implies a - b = \frac{\pi}{4} \quad \dots(1)$
For $f(x)$ to be continuous at $x = \frac{\pi}{2}$,the left-hand limit must equal the right-hand limit:
$\lim_{x \rightarrow \frac{\pi}{2}^-} (2x \cot x + b) = 2\left(\frac{\pi}{2}\right) \lim_{x \rightarrow \frac{\pi}{2}^-} \cot x + b = \pi(0) + b = b$
$\lim_{x \rightarrow \frac{\pi}{2}^+} (a \cos 2x - b \sin x) = a \cos(\pi) - b \sin\left(\frac{\pi}{2}\right) = -a - b$
Equating them: $b = -a - b \implies a + 2b = 0 \implies a = -2b \quad \dots(2)$
Substitute $(2)$ into $(1)$: $-2b - b = \frac{\pi}{4} \implies -3b = \frac{\pi}{4} \implies b = -\frac{\pi}{12}$
Then $a = -2(-\frac{\pi}{12}) = \frac{\pi}{6}$.

(D) For a function to be continuous at $x = 1$,the left-hand limit,right-hand limit,and the value of the function at $x = 1$ must be equal.
$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (3ax + b) = 3a + b$
$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (5ax - 2b) = 5a - 2b$
Given $f(1) = 11$,we have:
$3a + b = 11$ (Equation $1$)
$5a - 2b = 11$ (Equation $2$)
From Equation $1$,$b = 11 - 3a$. Substituting this into Equation $2$:
$5a - 2(11 - 3a) = 11$
$5a - 22 + 6a = 11$
$11a = 33 \implies a = 3$
Substituting $a = 3$ into $b = 11 - 3a$:
$b = 11 - 3(3) = 11 - 9 = 2$
Thus,$a = 3$ and $b = 2$.

(B) Given $f(x) = \frac{(1 + \cos x) - \sin x}{(1 + \cos x) + \sin x}$.
Since $f(x)$ is continuous at $x = \pi$,we have $f(\pi) = \lim_{x \rightarrow \pi} f(x)$.
Using trigonometric identities $1 + \cos x = 2 \cos^2 \frac{x}{2}$ and $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$:
$\lim_{x \rightarrow \pi} f(x) = \lim_{x \rightarrow \pi} \frac{2 \cos^2 \frac{x}{2} - 2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2} + 2 \sin \frac{x}{2} \cos \frac{x}{2}}$
$= \lim_{x \rightarrow \pi} \frac{2 \cos \frac{x}{2} (\cos \frac{x}{2} - \sin \frac{x}{2})}{2 \cos \frac{x}{2} (\cos \frac{x}{2} + \sin \frac{x}{2})}$
$= \lim_{x \rightarrow \pi} \frac{\cos \frac{x}{2} - \sin \frac{x}{2}}{\cos \frac{x}{2} + \sin \frac{x}{2}}$
Dividing numerator and denominator by $\cos \frac{x}{2}$:
$= \lim_{x \rightarrow \pi} \frac{1 - \tan \frac{x}{2}}{1 + \tan \frac{x}{2}} = \lim_{x \rightarrow \pi} \tan \left( \frac{\pi}{4} - \frac{x}{2} \right)$
$= \tan \left( \frac{\pi}{4} - \frac{\pi}{2} \right) = \tan \left( -\frac{\pi}{4} \right) = -1$.

(C) For $f(x)$ to be continuous in $[-\pi, \pi]$,it must be continuous at $x = -\pi/2$ and $x = \pi/2$.
At $x = -\pi/2$:
$\lim_{x \to -\pi/2^-} f(x) = -2 \sin(-\pi/2) = -2(-1) = 2$.
$\lim_{x \to -\pi/2^+} f(x) = a \sin(-\pi/2) + b = -a + b$.
Since it is continuous,$-a + b = 2$ (Equation $1$).
At $x = \pi/2$:
$\lim_{x \to \pi/2^-} f(x) = a \sin(\pi/2) + b = a + b$.
$\lim_{x \to \pi/2^+} f(x) = \cos(\pi/2) = 0$.
Since it is continuous,$a + b = 0$ (Equation $2$).
Adding Equation $1$ and Equation $2$: $(-a + b) + (a + b) = 2 + 0 \implies 2b = 2 \implies b = 1$.
Substituting $b = 1$ in Equation $2$: $a + 1 = 0 \implies a = -1$.
Now,calculate $(3a + 2b)^3$:
$(3(-1) + 2(1))^3 = (-3 + 2)^3 = (-1)^3 = -1$.

(C) Given $f(x) = \begin{cases} x, & x \le 0 \\ 0, & x > 0 \end{cases}$.
For continuity at $x = 0$:
Left Hand Limit $(LHL)$ = $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} x = 0$.
Right Hand Limit $(RHL)$ = $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} 0 = 0$.
Value of function $f(0) = 0$.
Since $LHL$ = $RHL$ = $f(0)$,the function is continuous at $x = 0$.
For differentiability at $x = 0$:
Left Hand Derivative $(LHD)$ = $\lim_{h \to 0^+} \frac{f(0-h) - f(0)}{-h} = \lim_{h \to 0^+} \frac{(0-h) - 0}{-h} = \lim_{h \to 0^+} \frac{-h}{-h} = 1$.
Right Hand Derivative $(RHD)$ = $\lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^+} \frac{0 - 0}{h} = 0$.
Since $LHD$ $\neq$ $RHD$,the function is not differentiable at $x = 0$.

(A) For $f(x)$ to be continuous at $x = -2$,the left-hand limit must equal the right-hand limit.
$\lim_{x \rightarrow -2^{-}} f(x) = \lim_{x \rightarrow -2^{-}} (6 \beta - 3 \alpha x) = 6 \beta - 3 \alpha (-2) = 6 \beta + 6 \alpha$.
$\lim_{x \rightarrow -2^{+}} f(x) = \lim_{x \rightarrow -2^{+}} (4x + 1) = 4(-2) + 1 = -8 + 1 = -7$.
Since $f(x)$ is continuous,$6 \beta + 6 \alpha = -7$.
Dividing both sides by $6$,we get $\alpha + \beta = \frac{-7}{6}$.

(C) Since $f(x)$ is continuous at $x = 0$,we have $f(0) = \lim_{x \to 0} f(x)$.
$f(0) = \lim_{x \to 0} \frac{(e^{2x} - 1) \sin x^{\circ}}{x^2}$
Using the conversion $x^{\circ} = \frac{x\pi}{180}$ radians,we get:
$f(0) = \lim_{x \to 0} \frac{(e^{2x} - 1)}{x} \cdot \frac{\sin(\frac{x\pi}{180})}{x}$
$f(0) = \lim_{x \to 0} \left( 2 \cdot \frac{e^{2x} - 1}{2x} \right) \cdot \left( \frac{\pi}{180} \cdot \frac{\sin(\frac{x\pi}{180})}{\frac{x\pi}{180}} \right)$
Using the standard limits $\lim_{u \to 0} \frac{e^u - 1}{u} = 1$ and $\lim_{v \to 0} \frac{\sin v}{v} = 1$:
$f(0) = 2(1) \cdot \frac{\pi}{180}(1) = \frac{2\pi}{180} = \frac{\pi}{90}$.

(D) For $f(x)$ to be continuous at $x = \frac{\pi}{2}$,we must have $\lim_{x \rightarrow \frac{\pi}{2}} f(x) = f(\frac{\pi}{2}) = k$.
First,simplify the expression for $f(x)$:
$f(x) = \frac{(1+\cos 2x) - \sin 2x}{(1+\cos 2x) + \sin 2x} = \frac{2\cos^2 x - 2\sin x \cos x}{2\cos^2 x + 2\sin x \cos x}$.
Factoring out $2\cos x$ from the numerator and denominator:
$f(x) = \frac{2\cos x(\cos x - \sin x)}{2\cos x(\cos x + \sin x)} = \frac{\cos x - \sin x}{\cos x + \sin x}$.
Now,evaluate the limit as $x \rightarrow \frac{\pi}{2}$:
$\lim_{x \rightarrow \frac{\pi}{2}} \frac{\cos x - \sin x}{\cos x + \sin x} = \frac{\cos(\frac{\pi}{2}) - \sin(\frac{\pi}{2})}{\cos(\frac{\pi}{2}) + \sin(\frac{\pi}{2})} = \frac{0 - 1}{0 + 1} = -1$.
Thus,$k = -1$.

(C) Since $f(x)$ is continuous at $x = 0$,we must have $\lim_{x \to 0} f(x) = f(0) = k$.
Thus,$k = \lim_{x \to 0} \left[\tan \left(\frac{\pi}{4} + x\right)\right]^{\frac{1}{x}}$.
Using the formula $\tan(\frac{\pi}{4} + x) = \frac{1 + \tan x}{1 - \tan x}$,we get:
$k = \lim_{x \to 0} \left(\frac{1 + \tan x}{1 - \tan x}\right)^{\frac{1}{x}}$.
This is of the form $1^{\infty}$,so we use the limit property $\lim_{x \to 0} (1 + u(x))^{v(x)} = e^{\lim_{x \to 0} u(x)v(x)}$.
$k = \exp \left[ \lim_{x \to 0} \frac{1}{x} \left( \frac{1 + \tan x}{1 - \tan x} - 1 \right) \right]$.
$k = \exp \left[ \lim_{x \to 0} \frac{1}{x} \left( \frac{1 + \tan x - 1 + \tan x}{1 - \tan x} \right) \right] = \exp \left[ \lim_{x \to 0} \frac{2 \tan x}{x(1 - \tan x)} \right]$.
Since $\lim_{x \to 0} \frac{\tan x}{x} = 1$,we have $k = e^{2(1)/(1-0)} = e^{2}$.

(A) For the function $f(x)$ to be continuous at $x = 0$,the limit of $f(x)$ as $x \rightarrow 0$ must equal $f(0)$.
$\lim _{x \rightarrow 0} f(x) = f(0) = 2$.
$\lim _{x \rightarrow 0} \frac{81^{x}-9^{x}}{k^{x}-1} = 2$.
Factor out $9^{x}$ from the numerator: $\lim _{x \rightarrow 0} \frac{9^{x}(9^{x}-1)}{k^{x}-1} = 2$.
Divide the numerator and denominator by $x$: $\lim _{x \rightarrow 0} \frac{9^{x} \cdot \frac{9^{x}-1}{x}}{\frac{k^{x}-1}{x}} = 2$.
Using the standard limit formula $\lim _{x \rightarrow 0} \frac{a^{x}-1}{x} = \ln a$,we get:
$\frac{9^{0} \cdot \ln 9}{\ln k} = 2$.
Since $9^{0} = 1$,we have $\frac{\ln 9}{\ln k} = 2$.
$\ln 9 = 2 \ln k \Rightarrow \ln 9 = \ln k^{2}$.
$k^{2} = 9 \Rightarrow k = 3$ (since $k$ must be positive for the base of an exponential function).
Thus,the value of $k$ is $3$.

(C) For the function $f(x)$ to be continuous at $x = 0$,we must have $\lim_{x \rightarrow 0} f(x) = f(0) = k$.
Given $f(x) = \frac{\log 10 + \log(0.1 + 2x)}{2x}$ for $x \neq 0$.
Using the property $\log a + \log b = \log(ab)$,we get:
$\log 10 + \log(0.1 + 2x) = \log(10 \times (0.1 + 2x)) = \log(1 + 20x)$.
So,$\lim_{x \rightarrow 0} \frac{\log(1 + 20x)}{2x} = k$.
We know that $\lim_{u \rightarrow 0} \frac{\log(1 + u)}{u} = 1$.
Multiplying and dividing by $20$,we get:
$\lim_{x \rightarrow 0} \frac{\log(1 + 20x)}{20x} \times 10 = k$.
Since $\lim_{x \rightarrow 0} \frac{\log(1 + 20x)}{20x} = 1$,we have $1 \times 10 = k$,so $k = 10$.
Therefore,$k + 2 = 10 + 2 = 12$.

(D) Given $f(x) = \frac{|x-2|}{x-2}$ for $x \neq 2$ and $f(2) = 1$.
We know that $|x-2| = (x-2)$ if $x > 2$ and $|x-2| = -(x-2)$ if $x < 2$.
For the right-hand limit: $\lim_{x \rightarrow 2^{+}} f(x) = \lim_{x \rightarrow 2^{+}} \frac{x-2}{x-2} = 1$.
For the left-hand limit: $\lim_{x \rightarrow 2^{-}} f(x) = \lim_{x \rightarrow 2^{-}} \frac{-(x-2)}{x-2} = -1$.
Since $\lim_{x \rightarrow 2^{+}} f(x) \neq \lim_{x \rightarrow 2^{-}} f(x)$,the limit does not exist at $x=2$.
Also,$f(2) = 1$. Since the limit does not exist,the function $f(x)$ is discontinuous at $x=2$.

(A) For the function $f(x)$ to be continuous at $x = 0$,we must have $f(0) = \lim_{x \rightarrow 0} f(x)$.
Calculating the limit: $\lim_{x \rightarrow 0} \left(\frac{1+4x}{1-4x}\right)^{\frac{1}{x}} = \frac{\lim_{x \rightarrow 0} (1+4x)^{\frac{1}{x}}}{\lim_{x \rightarrow 0} (1-4x)^{\frac{1}{x}}}$.
Using the standard limit formula $\lim_{u \rightarrow 0} (1+u)^{\frac{1}{u}} = e$,we rewrite the expression:
$= \frac{\left[\lim_{x \rightarrow 0} (1+4x)^{\frac{1}{4x}}\right]^{4}}{\left[\lim_{x \rightarrow 0} (1-4x)^{\frac{1}{-4x}}\right]^{-4}} = \frac{e^{4}}{e^{-4}}$.
$= e^{4 - (-4)} = e^{8}$.

(B) The function $f(x)$ is defined as:
$f(x) = \frac{1}{x-1}$ for $0 \leq x \leq 2$
$f(x) = \frac{x+5}{x+3}$ for $2 < x \leq 4$
Step $1$: Check for discontinuity within the intervals.
For $0 \leq x \leq 2$,the function $f(x) = \frac{1}{x-1}$ is undefined at $x=1$. Since $1 \in [0, 2]$,$x=1$ is a point of discontinuity.
For $2 < x \leq 4$,the function $f(x) = \frac{x+5}{x+3}$ is undefined at $x=-3$. Since $-3 \notin (2, 4]$,there are no points of discontinuity in this interval.
Step $2$: Check for continuity at the boundary point $x=2$.
$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} \frac{1}{x-1} = \frac{1}{2-1} = 1$
$\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} \frac{x+5}{x+3} = \frac{2+5}{2+3} = \frac{7}{5}$
Since $\lim_{x \to 2^-} f(x) \neq \lim_{x \to 2^+} f(x)$,the function is discontinuous at $x=2$.
Conclusion: The points of discontinuity are $x=1$ and $x=2$.

(A) Since $f(x)$ is continuous at $x = 0$,the limit of the function as $x \to 0$ must equal the value of the function at $x = 0$.
$\lim_{x \to 0} f(x) = f(0)$
$\lim_{x \to 0} \frac{4 \sin \pi x}{5 x} = 2k$
We know that $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$.
$\lim_{x \to 0} \left( \frac{4 \sin \pi x}{5 x} \right) = \lim_{x \to 0} \left( \frac{4 \pi \sin \pi x}{5 \pi x} \right) = \frac{4 \pi}{5} \lim_{x \to 0} \frac{\sin \pi x}{\pi x} = \frac{4 \pi}{5} \times 1 = \frac{4 \pi}{5}$.
Equating this to $f(0) = 2k$:
$\frac{4 \pi}{5} = 2k$
$k = \frac{4 \pi}{10} = \frac{2 \pi}{5}$
Note: Upon re-evaluating the calculation,the correct value is $\frac{2 \pi}{5}$,which corresponds to option $A$.

(A) Since $f$ is continuous at $x = \pi$,we have $f(\pi) = \lim_{x \rightarrow \pi} f(x)$.
Using $L$'$H$ôpital's rule for the limit $\lim_{x \rightarrow \pi} \frac{1 - \sin x + \cos x}{1 + \sin x + \cos x}$:
$\lim_{x \rightarrow \pi} \frac{\frac{d}{dx}(1 - \sin x + \cos x)}{\frac{d}{dx}(1 + \sin x + \cos x)} = \lim_{x \rightarrow \pi} \frac{-\cos x - \sin x}{\cos x - \sin x}$.
Substituting $x = \pi$:
$f(\pi) = \frac{-\cos(\pi) - \sin(\pi)}{\cos(\pi) - \sin(\pi)} = \frac{-(-1) - 0}{-1 - 0} = \frac{1}{-1} = -1$.

(A) To check continuity at $x=0$,we evaluate $\lim_{x \rightarrow 0} f(x)$.
Given $f(x) = \frac{(e^{3x}-1) \sin x^{\circ}}{x^2}$ for $x \neq 0$.
We know that $x^{\circ} = \frac{\pi x}{180}$ radians.
So,$\lim_{x \rightarrow 0} f(x) = \lim_{x \rightarrow 0} \frac{(e^{3x}-1) \sin(\frac{\pi x}{180})}{x^2}$.
Multiplying and dividing by $3x$ and $\frac{\pi}{180}$,we get:
$\lim_{x \rightarrow 0} \left( \frac{e^{3x}-1}{3x} \cdot 3 \right) \cdot \left( \frac{\sin(\frac{\pi x}{180})}{\frac{\pi x}{180}} \cdot \frac{\pi}{180} \right) \cdot \frac{1}{x} \cdot x^2$ is not correct,let us simplify:
$\lim_{x \rightarrow 0} \frac{e^{3x}-1}{x} \cdot \frac{\sin(\frac{\pi x}{180})}{x} = \lim_{x \rightarrow 0} \left( \frac{e^{3x}-1}{3x} \cdot 3 \right) \cdot \left( \frac{\sin(\frac{\pi x}{180})}{\frac{\pi x}{180}} \cdot \frac{\pi}{180} \right) = (1 \cdot 3) \cdot (1 \cdot \frac{\pi}{180}) = \frac{3\pi}{180} = \frac{\pi}{60}$.
Since $\lim_{x \rightarrow 0} f(x) = f(0) = \frac{\pi}{60}$,the function $f(x)$ is continuous at $x=0$.

(C) Given that $f(x)$ is continuous at $x = 3$.
For a function to be continuous at a point $x = c$,the left-hand limit,right-hand limit,and the value of the function at that point must be equal.
$\therefore \lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x) = f(3)$.
Calculating the left-hand limit: $\lim_{x \to 3^-} f(x) = \lim_{h \to 0} f(3 - h) = \lim_{h \to 0} [a(3 - h) + 1] = 3a + 1$.
Calculating the right-hand limit: $\lim_{x \to 3^+} f(x) = \lim_{h \to 0} f(3 + h) = \lim_{h \to 0} [b(3 + h) + 3] = 3b + 3$.
Since $f(x)$ is continuous at $x = 3$,we equate the limits: $3a + 1 = 3b + 3$.
Rearranging the terms: $3a - 3b = 3 - 1$.
$3(a - b) = 2$.
$a - b = \frac{2}{3}$.

(C) Given function is $f(x) = \begin{cases} x - \frac{|x|}{x}, & x < 0 \\ x + \frac{|x|}{x}, & x > 0 \\ 1, & x = 0 \end{cases}$.
To check continuity at $x = 0$,we evaluate the left-hand limit $(LHL)$,right-hand limit $(RHL)$,and the value of the function $f(0)$.
$LHL = \lim_{x \to 0^{-}} f(x) = \lim_{x \to 0^{-}} \left(x - \frac{|x|}{x}\right)$.
Since $x < 0$,$|x| = -x$,so $\frac{|x|}{x} = -1$.
$LHL = \lim_{x \to 0^{-}} (x - (-1)) = \lim_{x \to 0^{-}} (x + 1) = 0 + 1 = 1$.
$RHL = \lim_{x \to 0^{+}} f(x) = \lim_{x \to 0^{+}} \left(x + \frac{|x|}{x}\right)$.
Since $x > 0$,$|x| = x$,so $\frac{|x|}{x} = 1$.
$RHL = \lim_{x \to 0^{+}} (x + 1) = 0 + 1 = 1$.
Given $f(0) = 1$.
Since $LHL = RHL = f(0) = 1$,the function $f(x)$ is continuous at $x = 0$.

(D) Given that the function $f(x)$ is continuous at $x = 0$,we must have $\lim_{x \to 0} f(x) = f(0)$.
Since $f(0) = 16$,we evaluate the limit:
$\lim_{x \to 0} \frac{(e^{kx} - 1) \tan kx}{4x^2} = 16$.
We can rewrite the expression as:
$\lim_{x \to 0} \left( \frac{e^{kx} - 1}{kx} \right) \left( \frac{\tan kx}{kx} \right) \left( \frac{k^2 x^2}{4x^2} \right) = 16$.
Using the standard limits $\lim_{u \to 0} \frac{e^u - 1}{u} = 1$ and $\lim_{u \to 0} \frac{\tan u}{u} = 1$,we get:
$1 \times 1 \times \frac{k^2}{4} = 16$.
$\frac{k^2}{4} = 16$.
$k^2 = 64$.
$k = \pm 8$.

(B) Given that the function $f(x)$ is continuous at $x = 0$,we have $f(0) = \lim_{x \to 0} f(x)$.
$f(0) = \lim_{x \to 0} \frac{\log(1 + ax) - \log(1 - bx)}{x}$.
Since this is a $\frac{0}{0}$ indeterminate form,we apply $L'\text{Hospital's rule}$ by differentiating the numerator and denominator with respect to $x$:
$f(0) = \lim_{x \to 0} \frac{\frac{d}{dx}(\log(1 + ax)) - \frac{d}{dx}(\log(1 - bx))}{\frac{d}{dx}(x)}$.
$f(0) = \lim_{x \to 0} \frac{\frac{a}{1 + ax} - \frac{-b}{1 - bx}}{1}$.
$f(0) = \lim_{x \to 0} \left( \frac{a}{1 + ax} + \frac{b}{1 - bx} \right)$.
Substituting $x = 0$:
$f(0) = \frac{a}{1 + 0} + \frac{b}{1 - 0} = a + b$.

(C) function $f(x)$ is continuous at $x = a$ if $\lim_{x \to a} f(x) = f(a)$.
For option $C$,we check the limit at $x = 0$:
$R.H.L. = \lim_{x \to 0^+} \frac{e^{1/x} - 1}{e^{1/x} + 1} = \lim_{x \to 0^+} \frac{1 - e^{-1/x}}{1 + e^{-1/x}} = \frac{1 - 0}{1 + 0} = 1$.
$L.H.L. = \lim_{x \to 0^-} \frac{e^{1/x} - 1}{e^{1/x} + 1} = \frac{0 - 1}{0 + 1} = -1$.
Since $L.H.L. \neq R.H.L.$,the limit does not exist at $x = 0$.
Therefore,the function is not continuous at $x = 0$.

(D) Since $f(x)$ is continuous at $x = 0$,we have $f(0) = \lim_{x \to 0} f(x)$.
$f(0) = \lim_{x \to 0} \frac{e^{x^2} - \cos x}{x^2}$
To evaluate this limit,we add and subtract $1$ in the numerator:
$f(0) = \lim_{x \to 0} \frac{(e^{x^2} - 1) + (1 - \cos x)}{x^2}$
$f(0) = \lim_{x \to 0} \frac{e^{x^2} - 1}{x^2} + \lim_{x \to 0} \frac{1 - \cos x}{x^2}$
Using standard limits $\lim_{u \to 0} \frac{e^u - 1}{u} = 1$ and $\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$:
$f(0) = 1 + \frac{1}{2} = \frac{3}{2}$.

(C) For $f(x)$ to be continuous at $x = 0$,the left-hand limit must equal the right-hand limit and the function value at $x = 0$.
$\lim_{x \to 0^+} (x^2 + \alpha) = \lim_{x \to 0^-} (2\sqrt{x^2 + 1} + \beta)$
$0^2 + \alpha = 2\sqrt{0^2 + 1} + \beta$
$\alpha = 2 + \beta \implies \alpha - \beta = 2 . . . (1)$
Given $f(\frac{1}{2}) = 2$. Since $\frac{1}{2} \ge 0$,we use the first part of the function:
$f(\frac{1}{2}) = (\frac{1}{2})^2 + \alpha = 2$
$\frac{1}{4} + \alpha = 2$
$\alpha = 2 - \frac{1}{4} = \frac{7}{4}$
Substitute $\alpha = \frac{7}{4}$ into equation $(1)$:
$\frac{7}{4} - \beta = 2$
$\beta = \frac{7}{4} - 2 = -\frac{1}{4}$
Now,calculate $\alpha^2 + \beta^2$:
$\alpha^2 + \beta^2 = (\frac{7}{4})^2 + (-\frac{1}{4})^2$
$\alpha^2 + \beta^2 = \frac{49}{16} + \frac{1}{16} = \frac{50}{16} = \frac{25}{8}$

(B) Since $f(x)$ is continuous at $x = 0$,we have $f(0) = \lim_{x \to 0} f(x)$.
$K = \lim_{x \to 0} \log(\sec^2 x)^{\cot^2 x}$
Using the property $\log(a^b) = b \log a$,we get:
$K = \lim_{x \to 0} \cot^2 x \cdot \log(\sec^2 x)$
Since $\sec^2 x = 1 + \tan^2 x$,we have:
$K = \lim_{x \to 0} \cot^2 x \cdot \log(1 + \tan^2 x)$
$K = \lim_{x \to 0} \frac{\log(1 + \tan^2 x)}{\tan^2 x}$
Using the standard limit $\lim_{u \to 0} \frac{\log(1+u)}{u} = 1$,where $u = \tan^2 x$ as $x \to 0$:
$K = 1$.

(C) Since $f(x)$ is continuous at $x = 0$,we have $f(0) = \lim_{x \to 0} f(x)$.
Given $f(0) = K$,so $K = \lim_{x \to 0} [\tan(\frac{\pi}{4} + x)]^{\frac{1}{x}}$.
Using the formula $\tan(\frac{\pi}{4} + x) = \frac{1 + \tan x}{1 - \tan x}$,we get:
$K = \lim_{x \to 0} (\frac{1 + \tan x}{1 - \tan x})^{\frac{1}{x}}$.
This is of the form $1^{\infty}$,so we use the formula $\lim_{x \to 0} (1 + g(x))^{h(x)} = e^{\lim_{x \to 0} g(x)h(x)}$.
$K = \lim_{x \to 0} [1 + (\frac{1 + \tan x}{1 - \tan x} - 1)]^{\frac{1}{x}} = \lim_{x \to 0} [1 + \frac{2 \tan x}{1 - \tan x}]^{\frac{1}{x}}$.
$K = e^{\lim_{x \to 0} (\frac{2 \tan x}{1 - \tan x} \cdot \frac{1}{x})}$.
Since $\lim_{x \to 0} \frac{\tan x}{x} = 1$,we have:
$K = e^{\lim_{x \to 0} (\frac{2}{1 - \tan x} \cdot 1)} = e^{\frac{2}{1 - 0}} = e^{2}$.

(C) Since $f(x)$ is continuous at $x = 0$,we must have $\lim_{x \to 0} f(x) = f(0)$.
Therefore,$\lim_{x \to 0} \frac{\text{log}(1 + 2x) \cdot \sin x^{\circ}}{x^2} = k$.
We know that $\sin x^{\circ} = \sin(\frac{\pi x}{180})$.
So,$\lim_{x \to 0} \frac{\text{log}(1 + 2x)}{x} \cdot \frac{\sin(\frac{\pi x}{180})}{x} = k$.
Multiplying and dividing by $2$ and $\frac{\pi}{180}$ respectively:
$\lim_{x \to 0} \left( 2 \cdot \frac{\text{log}(1 + 2x)}{2x} \right) \cdot \left( \frac{\pi}{180} \cdot \frac{\sin(\frac{\pi x}{180})}{\frac{\pi x}{180}} \right) = k$.
Using the standard limits $\lim_{u \to 0} \frac{\text{log}(1+u)}{u} = 1$ and $\lim_{v \to 0} \frac{\sin v}{v} = 1$:
$2 \cdot 1 \cdot \frac{\pi}{180} \cdot 1 = k$.
Thus,$k = \frac{\pi}{90}$.

(A) Given the function $f(x) = \begin{cases} x \sin \frac{1}{x}, & x \neq 0 \\ k, & x = 0 \end{cases}$.
Since $f(x)$ is continuous at $x = 0$,the condition $\lim_{x \to 0} f(x) = f(0)$ must hold.
We calculate the limit: $\lim_{x \to 0} f(x) = \lim_{x \to 0} x \sin \frac{1}{x}$.
We know that for all $x \neq 0$,$-1 \leq \sin \frac{1}{x} \leq 1$.
Multiplying by $x$,we get $-|x| \leq x \sin \frac{1}{x} \leq |x|$.
By the Squeeze Theorem,as $x \to 0$,both $-|x|$ and $|x|$ approach $0$.
Therefore,$\lim_{x \to 0} x \sin \frac{1}{x} = 0$.
Since $f(0) = k$,we have $k = 0$.

(C) Given,$f(x) = \begin{cases} x \sin \left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x = 0 \end{cases}$
For continuity at $x = 0$:
$LHL = f(0-0) = \lim_{h \to 0} f(0-h) = \lim_{h \to 0} (-h) \sin \left(-\frac{1}{h}\right) = \lim_{h \to 0} h \sin \left(\frac{1}{h}\right) = 0 \times (\text{finite quantity}) = 0$.
$RHL = f(0+0) = \lim_{h \to 0} f(0+h) = \lim_{h \to 0} h \sin \left(\frac{1}{h}\right) = 0 \times (\text{finite quantity}) = 0$.
Since $f(0) = LHL = RHL = 0$,$f(x)$ is continuous at $x = 0$.
For differentiability at $x = 0$:
$Rf'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0} \frac{h \sin(1/h) - 0}{h} = \lim_{h \to 0} \sin(1/h)$.
As $h \to 0$,$\sin(1/h)$ oscillates between $-1$ and $1$,so the limit does not exist.
Similarly,$Lf'(0) = \lim_{h \to 0} \frac{f(0-h) - f(0)}{-h} = \lim_{h \to 0} \frac{-h \sin(-1/h) - 0}{-h} = \lim_{h \to 0} \sin(1/h)$,which also does not exist.
Since the derivative does not exist,$f(x)$ is not differentiable at $x = 0$.
Thus,$f(x)$ is continuous but not differentiable at $x = 0$.

(D) Given that $f(x)$ is continuous at $x = 0$,we have $f(0) = \lim_{x \to 0} f(x)$.
$f(0) = k$.
$\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\log (1 + 2ax) - \log (1 - bx)}{x}$.
This is a $\frac{0}{0}$ form,so we apply $L$'Hospital's rule:
$k = \lim_{x \to 0} \frac{\frac{d}{dx}(\log (1 + 2ax) - \log (1 - bx))}{\frac{d}{dx}(x)}$.
$k = \lim_{x \to 0} \frac{\frac{2a}{1 + 2ax} - \frac{-b}{1 - bx}}{1}$.
$k = \frac{2a}{1 + 0} + \frac{b}{1 - 0} = 2a + b$.

(C) For $f(x)$ to be continuous at $x = 0$,we must have $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$.
First,calculate the left-hand limit:
$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (1+|\sin x|)^{\frac{a}{|\sin x|}}$.
Let $u = |\sin x|$. As $x \to 0$,$u \to 0^+$. The limit becomes $\lim_{u \to 0^+} (1+u)^{a/u} = e^a$.
Next,calculate the right-hand limit:
$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} e^{\frac{\tan 2x}{\tan 3x}}$.
Using the limit $\lim_{\theta \to 0} \frac{\tan k\theta}{\theta} = k$,we get $\lim_{x \to 0^+} \frac{\tan 2x}{\tan 3x} = \lim_{x \to 0^+} \frac{(\tan 2x / 2x) \cdot 2x}{(\tan 3x / 3x) \cdot 3x} = \frac{1 \cdot 2}{1 \cdot 3} = \frac{2}{3}$.
Thus,$\lim_{x \to 0^+} f(x) = e^{2/3}$.
Since $f(0) = b$,for continuity,$e^a = e^{2/3} = b$.
Therefore,$a = 2/3$ and $b = e^{2/3}$.

(D) For the function $f(x)$ to be continuous in $[-\pi, \pi]$,it must be continuous at the transition points $x = -\frac{\pi}{2}$ and $x = \frac{\pi}{2}$.
At $x = -\frac{\pi}{2}$:
$\lim_{x \rightarrow -\frac{\pi}{2}^-} (-2 \sin x) = -2 \sin(-\frac{\pi}{2}) = -2(-1) = 2$.
$\lim_{x \rightarrow -\frac{\pi}{2}^+} (a \sin x + b) = a \sin(-\frac{\pi}{2}) + b = -a + b$.
Equating these,we get $-a + b = 2$ (Equation $1$).
At $x = \frac{\pi}{2}$:
$\lim_{x \rightarrow \frac{\pi}{2}^-} (a \sin x + b) = a \sin(\frac{\pi}{2}) + b = a + b$.
$\lim_{x \rightarrow \frac{\pi}{2}^+} \cos x = \cos(\frac{\pi}{2}) = 0$.
Equating these,we get $a + b = 0$ (Equation $2$).
Adding Equation $1$ and Equation $2$:
$(-a + b) + (a + b) = 2 + 0 \Rightarrow 2b = 2 \Rightarrow b = 1$.
Substituting $b = 1$ into Equation $2$:
$a + 1 = 0 \Rightarrow a = -1$.
Thus,the values are $a = -1$ and $b = 1$.

(C) For a function $f(x)$ to be continuous at $x = 0$,the condition $f(0) = \lim_{x \to 0} f(x)$ must be satisfied.
Given $f(0) = k$.
Now,we calculate the limit: $\lim_{x \to 0} f(x) = \lim_{x \to 0} x \sin \frac{1}{x}$.
We know that for all $x \neq 0$,$-1 \leq \sin \frac{1}{x} \leq 1$.
Multiplying by $x$ (where $x > 0$),we get $-x \leq x \sin \frac{1}{x} \leq x$.
By the Squeeze Theorem,as $x \to 0$,both $-x$ and $x$ approach $0$.
Therefore,$\lim_{x \to 0} x \sin \frac{1}{x} = 0$.
Since $f(0) = \lim_{x \to 0} f(x)$,we have $k = 0$.

(D) The function $f(x) = \frac{x+1}{9x+x^3}$ can be written as $f(x) = \frac{x+1}{x(9+x^2)}$.
$A$ function is discontinuous where the denominator is equal to zero,i.e.,$x(9+x^2) = 0$.
For this equation,we get $x = 0$ or $9+x^2 = 0$.
The equation $x^2 = -9$ has no real solutions because $x^2$ is always non-negative for all real $x$.
Therefore,the function is discontinuous only at $x = 0$.
Thus,the function is discontinuous at exactly one point.

(C) The function is defined as $f(x) = \begin{cases} 1 & x > 0 \\ -1 & x < 0 \\ 1 & x = 0 \end{cases}$.
To check for continuity at $x = 0$,we evaluate the left-hand limit and right-hand limit:
$\lim_{x \to 0^-} f(x) = -1$
$\lim_{x \to 0^+} f(x) = 1$
Since the left-hand limit is not equal to the right-hand limit,the function is discontinuous at $x = 0$.
Since the function is not continuous at $x = 0$,it is also not differentiable at $x = 0$.
Therefore,the function is neither continuous nor differentiable at $x = 0$.

(A) For $g(x)$ to be continuous at $x=3$,we must have $K = \lim_{x \to 3} g(x) = \lim_{x \to 3} \int_3^{f(x)} \frac{3t^2}{x-3} dt$.
Since the integral is of the form $\frac{0}{0}$ as $x \to 3$ (because $f(3)=3$),we apply $L$'$H$ôpital's Rule and Leibniz's Rule for differentiation under the integral sign.
Let $I(x) = \int_3^{f(x)} 3t^2 dt$. Then $g(x) = \frac{I(x)}{x-3}$.
By $L$'$H$ôpital's Rule,$K = \lim_{x \to 3} \frac{\frac{d}{dx} \int_3^{f(x)} 3t^2 dt}{\frac{d}{dx} (x-3)}$.
Using Leibniz's Rule,$\frac{d}{dx} \int_3^{f(x)} 3t^2 dt = 3(f(x))^2 \cdot f^{\prime}(x)$.
Thus,$K = \lim_{x \to 3} \frac{3(f(x))^2 \cdot f^{\prime}(x)}{1} = 3(f(3))^2 \cdot f^{\prime}(3)$.
Substituting the given values $f(3)=3$ and $f^{\prime}(3)=\frac{1}{27}$:
$K = 3 \cdot (3)^2 \cdot \frac{1}{27} = 3 \cdot 9 \cdot \frac{1}{27} = 27 \cdot \frac{1}{27} = 1$.

(C) The function is defined as $f(x) = [x]$.
For any integer $n$,the floor function $[x]$ is discontinuous at $x = n$ because the left-hand limit is $n-1$ and the right-hand limit is $n$.
Given the domain $x \in (-1, 2)$,the integers within this interval are $0$ and $1$.
Therefore,the function $f(x) = [x]$ is discontinuous at $x = 0$ and $x = 1$.

(C) For $f(x)$ to be continuous at $x = 0$,we must have $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^-} f(x) = f(0)$.
First,calculate the right-hand limit:
$\lim_{x \to 0^+} \frac{8^x-4^x-2^x+1}{x^2} = \lim_{x \to 0^+} \frac{(4^x-1)(2^x-1)}{x^2} = \lim_{x \to 0^+} \left( \frac{4^x-1}{x} \right) \left( \frac{2^x-1}{x} \right) = \log 4 \cdot \log 2$.
Next,calculate the left-hand limit and $f(0)$:
$f(0) = e^0 \sin(0) + 0 + \lambda \log 4 = \lambda \log 4$.
Equating the two limits:
$\log 4 \cdot \log 2 = \lambda \log 4 \implies \lambda = \log 2$.
Now,calculate $1000 e^\lambda$:
$1000 e^{\log 2} = 1000 \times 2 = 2000$.

(B) For a function $f(x)$ to be continuous at $x = 0$,the left-hand limit $(LHL)$,the right-hand limit $(RHL)$,and the value of the function $f(0)$ must be equal.
$LHL = \lim_{x \to 0^-} \frac{1-\cos 4x}{x^2} = \lim_{x \to 0^-} \frac{2\sin^2(2x)}{x^2} = \lim_{x \to 0^-} 2 \left(\frac{\sin 2x}{2x}\right)^2 \times 4 = 2 \times 1^2 \times 4 = 8$.
$RHL = \lim_{x \to 0^+} \frac{\sqrt{16+\sqrt{x}}-4}{\sqrt{x}}$.
Multiplying by the conjugate: $\lim_{x \to 0^+} \frac{(\sqrt{16+\sqrt{x}}-4)(\sqrt{16+\sqrt{x}}+4)}{\sqrt{x}(\sqrt{16+\sqrt{x}}+4)} = \lim_{x \to 0^+} \frac{16+\sqrt{x}-16}{\sqrt{x}(\sqrt{16+\sqrt{x}}+4)} = \lim_{x \to 0^+} \frac{1}{\sqrt{16+\sqrt{x}}+4} = \frac{1}{4+4} = \frac{1}{8}$.
Since $LHL \neq RHL$,the function is not continuous at $x = 0$ for any value of $a$. The problem statement contains a contradiction as the limits do not match.

(B) For the function to be continuous at $x = 0$,we must have $\lim_{x \to 0} f(x) = f(0) = -1$.
Evaluating the limit: $\lim_{x \to 0} \frac{\cos ax - \cos bx}{\cos cx - \cos bx}$.
Using the expansion $\cos \theta \approx 1 - \frac{\theta^2}{2}$ for small $\theta$:
$\lim_{x \to 0} \frac{(1 - \frac{a^2x^2}{2}) - (1 - \frac{b^2x^2}{2})}{(1 - \frac{c^2x^2}{2}) - (1 - \frac{b^2x^2}{2})} = \lim_{x \to 0} \frac{\frac{b^2x^2}{2} - \frac{a^2x^2}{2}}{\frac{b^2x^2}{2} - \frac{c^2x^2}{2}} = \frac{b^2 - a^2}{b^2 - c^2}$.
Setting this equal to $-1$: $\frac{b^2 - a^2}{b^2 - c^2} = -1$.
$b^2 - a^2 = -(b^2 - c^2) \implies b^2 - a^2 = c^2 - b^2$.
$2b^2 = a^2 + c^2$.
This condition implies that $a^2, b^2, c^2$ are in Arithmetic progression.

(B) We evaluate the left-hand limit at $x = 4$: $\lim_{x \rightarrow 4^{-}} f(x) = [4^{-}] - [\frac{4^{-}}{4}] = 3 - 0 = 3$.
We evaluate the right-hand limit at $x = 4$: $\lim_{x \rightarrow 4^{+}} f(x) = [4^{+}] - [\frac{4^{+}}{4}] = 4 - 1 = 3$.
We evaluate the function value at $x = 4$: $f(4) = [4] - [\frac{4}{4}] = 4 - 1 = 3$.
Since $\lim_{x \rightarrow 4^{-}} f(x) = \lim_{x \rightarrow 4^{+}} f(x) = f(4) = 3$,the function $f(x)$ is continuous at $x = 4$.

(C) For $f(x)$ to be continuous at $x=0$, $f(0) = \lim_{x \to 0} f(x)$.
Let $L = \lim_{x \to 0} \frac{(27-2x)^{1/3}-3}{9-3(243+5x)^{1/5}}$.
Factor out constants: $L = \lim_{x \to 0} \frac{3((1 - \frac{2x}{27})^{1/3} - 1)}{9(1 - (1 + \frac{5x}{243})^{1/5})} = \frac{1}{3} \lim_{x \to 0} \frac{(1 - \frac{2x}{27})^{1/3} - 1}{1 - (1 + \frac{5x}{243})^{1/5}}$.
Using the binomial expansion $(1+u)^n \approx 1+nu$ for small $u$:
Numerator: $(1 - \frac{2x}{27})^{1/3} - 1 \approx 1 - \frac{2x}{81} - 1 = -\frac{2x}{81}$.
Denominator: $1 - (1 + \frac{5x}{243})^{1/5} \approx 1 - (1 + \frac{x}{243}) = -\frac{x}{243}$.
Thus, $L = \frac{1}{3} \cdot \frac{-2x/81}{-x/243} = \frac{1}{3} \cdot \frac{2}{81} \cdot 243 = \frac{1}{3} \cdot 2 \cdot 3 = 2$.

(A) For $f(x)$ to be continuous at $x = 0$,$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^-} f(x) = f(0)$.
First,calculate the right-hand limit: $\lim_{x \to 0^+} \frac{8^x - 4^x - 2^x + 1}{x^2} = \lim_{x \to 0^+} \frac{(4^x - 1)(2^x - 1)}{x^2} = \lim_{x \to 0^+} \left( \frac{4^x - 1}{x} \right) \left( \frac{2^x - 1}{x} \right) = \log 4 \cdot \log 2$.
Next,calculate the left-hand limit: $\lim_{x \to 0^-} (e^x \sin x + kx + \lambda \log 4) = e^0 \sin 0 + k(0) + \lambda \log 4 = \lambda \log 4$.
Equating the two: $\lambda \log 4 = \log 4 \cdot \log 2 \implies \lambda = \log 2$.
Then $e^\lambda = e^{\log 2} = 2$.
Finally,$500 e^\lambda = 500 \times 2 = 1000$.

(B) Since $f(x)$ is continuous at $x = 0$,$f(0) = \lim_{x \to 0} f(x)$.
$f(0) = \lim_{x \to 0} \frac{10^x + 7^x - 14^x - 5^x}{1 - \cos x}$.
Factor the numerator: $10^x - 14^x + 7^x - 5^x = 2^x(5^x - 7^x) - 1(5^x - 7^x) = (2^x - 1)(5^x - 7^x)$.
So,$f(0) = \lim_{x \to 0} \frac{(2^x - 1)(5^x - 7^x)}{1 - \cos x}$.
Using the standard limits $\lim_{x \to 0} \frac{a^x - 1}{x} = \ln a$ and $\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$:
$f(0) = \lim_{x \to 0} \frac{\frac{2^x - 1}{x} \cdot \frac{5^x - 7^x}{x}}{\frac{1 - \cos x}{x^2}} = \frac{(\ln 2)(\ln 5 - \ln 7)}{1/2} = 2 \ln 2 \ln(5/7) = \ln 4 \ln(5/7)$.

(B) For $f(x)$ to be continuous at $x = 0$,we must have $f(0) = \lim_{x \to 0} f(x)$.
Given $f(x) = \frac{\sin(\pi \cos^2 x)}{3x^2}$.
Using the identity $\cos^2 x = 1 - \sin^2 x$,we have:
$f(x) = \frac{\sin(\pi(1 - \sin^2 x))}{3x^2} = \frac{\sin(\pi - \pi \sin^2 x)}{3x^2}$.
Since $\sin(\pi - \theta) = \sin \theta$,we get:
$f(x) = \frac{\sin(\pi \sin^2 x)}{3x^2}$.
Now,$\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\sin(\pi \sin^2 x)}{3x^2} = \lim_{x \to 0} \left( \frac{\sin(\pi \sin^2 x)}{\pi \sin^2 x} \times \frac{\pi \sin^2 x}{3x^2} \right)$.
Since $\lim_{u \to 0} \frac{\sin u}{u} = 1$ and $\lim_{x \to 0} \frac{\sin x}{x} = 1$,we have:
$f(0) = 1 \times \frac{\pi}{3} \times (1)^2 = \frac{\pi}{3}$.

(A) For $f(x)$ to be continuous at $x = 0$,we must have $\lim_{x \to 0} f(x) = f(0)$.
First,evaluate the limit: $\lim_{x \to 0} \frac{9^x - 2 \cdot 3^x + 1}{\log(1 + 3x) \cdot \tan 2x}$.
Note that $9^x - 2 \cdot 3^x + 1 = (3^x - 1)^2$.
So,the limit is $\lim_{x \to 0} \frac{(3^x - 1)^2}{\log(1 + 3x) \cdot \tan 2x}$.
Using standard limits $\lim_{x \to 0} \frac{3^x - 1}{x} = \log 3$,$\lim_{x \to 0} \frac{\log(1 + 3x)}{3x} = 1$,and $\lim_{x \to 0} \frac{\tan 2x}{2x} = 1$:
Limit $= \lim_{x \to 0} \frac{(\frac{3^x - 1}{x})^2 \cdot x^2}{(\frac{\log(1 + 3x)}{3x} \cdot 3x) \cdot (\frac{\tan 2x}{2x} \cdot 2x)} = \frac{(\log 3)^2}{3 \cdot 2} = \frac{(\log 3)^2}{6}$.
Given $f(0) = a(\log b)^c$,we have $a(\log b)^c = \frac{1}{6}(\log 3)^2$.
Comparing terms,we get $a = \frac{1}{6}$,$b = 3$,and $c = 2$.
Therefore,$a + b + c = \frac{1}{6} + 3 + 2 = \frac{1}{6} + 5 = \frac{31}{6}$.

(C) For $f(x)$ to be continuous at $x=1$,$f(1) = \lim_{x \to 1} f(x)$.
Let $1-x = h$. As $x \to 1$,$h \to 0$.
$f(1) = \lim_{h \to 0} \frac{1+\cos(\pi(1-h))}{\pi h^2}$
$f(1) = \lim_{h \to 0} \frac{1+\cos(\pi - \pi h)}{\pi h^2}$
Since $\cos(\pi - \theta) = -\cos \theta$,we have:
$f(1) = \lim_{h \to 0} \frac{1-\cos(\pi h)}{\pi h^2}$
Using the identity $1-\cos \theta = 2\sin^2(\frac{\theta}{2})$:
$f(1) = \lim_{h \to 0} \frac{2\sin^2(\frac{\pi h}{2})}{\pi h^2}$
$f(1) = \frac{2}{\pi} \lim_{h \to 0} \left( \frac{\sin(\frac{\pi h}{2})}{h} \right)^2$
Multiply and divide by $(\frac{\pi}{2})^2$:
$f(1) = \frac{2}{\pi} \cdot (\frac{\pi}{2})^2 \lim_{h \to 0} \left( \frac{\sin(\frac{\pi h}{2})}{\frac{\pi h}{2}} \right)^2$
$f(1) = \frac{2}{\pi} \cdot \frac{\pi^2}{4} \cdot (1)^2 = \frac{\pi}{2}$.

(A) Given that $f(x)$ is continuous at $x = 0$,we have $\lim_{x \to 0} f(x) = f(0) = k$.
Since the limit is of the form $1^\infty$,we use the formula $\lim_{x \to a} [g(x)]^{h(x)} = e^{\lim_{x \to a} [g(x) - 1]h(x)}$.
Here,$g(x) = \tan(\frac{\pi}{4} + x)$ and $h(x) = \frac{1}{x}$.
$k = e^{\lim_{x \to 0} [\tan(\frac{\pi}{4} + x) - 1] \cdot \frac{1}{x}}$.
Using $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,we get $\tan(\frac{\pi}{4} + x) = \frac{1 + \tan x}{1 - \tan x}$.
$k = e^{\lim_{x \to 0} [\frac{1 + \tan x}{1 - \tan x} - 1] \cdot \frac{1}{x}} = e^{\lim_{x \to 0} [\frac{1 + \tan x - 1 + \tan x}{1 - \tan x}] \cdot \frac{1}{x}}$.
$k = e^{\lim_{x \to 0} \frac{2 \tan x}{x(1 - \tan x)}} = e^{2 \cdot \lim_{x \to 0} \frac{\tan x}{x} \cdot \lim_{x \to 0} \frac{1}{1 - \tan x}}$.
Since $\lim_{x \to 0} \frac{\tan x}{x} = 1$ and $\lim_{x \to 0} \frac{1}{1 - \tan x} = 1$,we have $k = e^{2 \cdot 1 \cdot 1} = e^2$.

(B) For $f(x)$ to be continuous at $x = 0$,we must have $f(0) = \lim_{x \to 0} f(x)$.
$f(0) = \lim_{x \to 0} \frac{2^{x}-1}{1-3^{x}}$
$f(0) = \lim_{x \to 0} \frac{2^{x}-1}{-(3^{x}-1)}$
$f(0) = -\frac{\lim_{x \to 0} \frac{2^{x}-1}{x}}{\lim_{x \to 0} \frac{3^{x}-1}{x}}$
Using the standard limit $\lim_{x \to 0} \frac{a^{x}-1}{x} = \log a$,we get:
$f(0) = -\frac{\log 2}{\log 3}$

(B) For $f(x)$ to be continuous at $x=0$,we must have $f(0) = \lim_{x \to 0} f(x)$.
$\lim_{x \to 0} \left(\frac{1+\tan x}{1+\sin x}\right)^{\operatorname{cosec} x} = \lim_{x \to 0} \exp\left(\operatorname{cosec} x \cdot \ln\left(\frac{1+\tan x}{1+\sin x}\right)\right)$.
Using the expansion $\tan x \approx x + \frac{x^3}{3}$ and $\sin x \approx x - \frac{x^3}{6}$,we have:
$\frac{1+\tan x}{1+\sin x} \approx \frac{1+x}{1+x} = 1$.
Since the limit is of the form $1^\infty$,we use $\lim_{x \to 0} (1+u)^v = e^{\lim_{x \to 0} uv}$.
$\lim_{x \to 0} \frac{1}{\sin x} \left(\frac{1+\tan x}{1+\sin x} - 1\right) = \lim_{x \to 0} \frac{1}{\sin x} \left(\frac{\tan x - \sin x}{1+\sin x}\right) = \lim_{x \to 0} \frac{\tan x - \sin x}{\sin x(1+\sin x)}$.
Using $\tan x - \sin x = \sin x(\sec x - 1) \approx x \cdot \frac{x^2}{2} = \frac{x^3}{2}$ and $\sin x(1+\sin x) \approx x(1+x) \approx x$,the limit is $\lim_{x \to 0} \frac{x^3/2}{x} = 0$.
Thus,$f(0) = e^0 = 1$.