Concept of limits, Evaluation of algebric limits Questions in English

(C) To find the limit $\mathop {\lim }\limits_{x \to 3} [x(x+1)]$,we substitute $x = 3$ directly into the expression because it is a polynomial function.
$\mathop {\lim }\limits_{x \to 3} [x(x+1)] = 3(3+1) = 3(4) = 12$

(B) The given expression is a polynomial function $f(x) = 1+x+x^{2}+ \dots +x^{10}$.
Since polynomial functions are continuous everywhere,the limit as $x \to -1$ is simply the value of the function at $x = -1$.
Substituting $x = -1$ into the expression:
$f(-1) = 1 + (-1) + (-1)^{2} + (-1)^{3} + (-1)^{4} + (-1)^{5} + (-1)^{6} + (-1)^{7} + (-1)^{8} + (-1)^{9} + (-1)^{10}$
$f(-1) = 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1$
Grouping the terms: $(1-1) + (1-1) + (1-1) + (1-1) + (1-1) + 1$
$f(-1) = 0 + 0 + 0 + 0 + 0 + 1 = 1$
Therefore,the limit is $1$.

(A) The given function is a rational function. Since the denominator is not zero at $x = 1$,we can evaluate the limit by direct substitution.
$\mathop {\lim }\limits_{x \to 1} \frac{x^{2}+1}{x+100} = \frac{1^{2}+1}{1+100}$
$= \frac{1+1}{101}$
$= \frac{2}{101}$

(A) The given function is a rational function. We first evaluate the function at $x = 2$. Substituting $x = 2$ gives $\frac{2^3 - 4(2^2) + 4(2)}{2^2 - 4} = \frac{8 - 16 + 8}{4 - 4} = \frac{0}{0}$.
Since it is an indeterminate form,we simplify the expression by factoring the numerator and the denominator.
Numerator: $x^3 - 4x^2 + 4x = x(x^2 - 4x + 4) = x(x - 2)^2$.
Denominator: $x^2 - 4 = (x - 2)(x + 2)$.
Now,$\mathop {\lim }\limits_{x \to 2} \frac{x(x-2)^2}{(x-2)(x+2)} = \mathop {\lim }\limits_{x \to 2} \frac{x(x-2)}{x+2}$ (for $x \neq 2$).
Substituting $x = 2$ in the simplified expression: $\frac{2(2-2)}{2+2} = \frac{2(0)}{4} = \frac{0}{4} = 0$.

(D) We are given the limit $\mathop {\lim }\limits_{x \to 2} \frac{x^{2}-4}{x^{3}-4 x^{2}+4 x}$.
First,we evaluate the function at $x = 2$:
$\frac{2^{2}-4}{2^{3}-4(2)^{2}+4(2)} = \frac{4-4}{8-16+8} = \frac{0}{0}$.
Since it is an indeterminate form,we simplify the expression by factoring:
$\frac{x^{2}-4}{x(x^{2}-4x+4)} = \frac{(x-2)(x+2)}{x(x-2)^{2}}$.
Canceling the common factor $(x-2)$,we get:
$\mathop {\lim }\limits_{x \to 2} \frac{x+2}{x(x-2)}$.
As $x \to 2$,the numerator approaches $4$ and the denominator approaches $2(0) = 0$.
Since the limit results in a non-zero constant divided by zero,the limit does not exist (it approaches infinity).

(A) The given function is a rational function. We first evaluate the function at $x = 2$. Substituting $x = 2$ in the expression,we get $\frac{2^3 - 2(2)^2}{2^2 - 5(2) + 6} = \frac{8 - 8}{4 - 10 + 6} = \frac{0}{0}$.
Since the form is $\frac{0}{0}$,we simplify the expression by factoring the numerator and the denominator.
$\mathop {\lim }\limits_{x \to 2} \frac{x^2(x-2)}{(x-2)(x-3)}$
Canceling the common factor $(x-2)$ for $x \neq 2$,we get:
$\mathop {\lim }\limits_{x \to 2} \frac{x^2}{x-3}$
Now,substituting $x = 2$:
$\frac{2^2}{2-3} = \frac{4}{-1} = -4$.

(A) We are given the limit $\mathop {\lim }\limits_{x \to 1} \left[\frac{x-2}{x^{2}-x}-\frac{1}{x^{3}-3 x^{2}+2 x}\right]$.
First,we factorize the denominators:
$x^2 - x = x(x-1)$
$x^3 - 3x^2 + 2x = x(x^2 - 3x + 2) = x(x-1)(x-2)$
Now,we rewrite the expression:
$\frac{x-2}{x(x-1)} - \frac{1}{x(x-1)(x-2)} = \frac{(x-2)^2 - 1}{x(x-1)(x-2)}$
$= \frac{x^2 - 4x + 4 - 1}{x(x-1)(x-2)} = \frac{x^2 - 4x + 3}{x(x-1)(x-2)}$
$= \frac{(x-3)(x-1)}{x(x-1)(x-2)}$
Canceling the common factor $(x-1)$ for $x \neq 1$:
$= \frac{x-3}{x(x-2)}$
Now,evaluating the limit as $x \to 1$:
$\mathop {\lim }\limits_{x \to 1} \frac{x-3}{x(x-2)} = \frac{1-3}{1(1-2)} = \frac{-2}{-1} = 2$.

(B) We use the standard limit formula: $\mathop {\lim }\limits_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}$.
Given expression: $\mathop {\lim }\limits_{x \to 1} \frac{x^{15}-1}{x^{10}-1}$.
Divide the numerator and denominator by $(x-1)$:
$= \frac{\mathop {\lim }\limits_{x \to 1} \frac{x^{15}-1}{x-1}}{\mathop {\lim }\limits_{x \to 1} \frac{x^{10}-1}{x-1}}$.
Applying the formula where $a=1$:
$= \frac{15(1)^{14}}{10(1)^9} = \frac{15}{10} = \frac{3}{2}$.

(A) To evaluate the limit $\mathop {\lim }\limits_{x \to 0} \frac{\sqrt{1+x}-1}{x}$,we can rationalize the numerator:
$\mathop {\lim }\limits_{x \to 0} \frac{\sqrt{1+x}-1}{x} \times \frac{\sqrt{1+x}+1}{\sqrt{1+x}+1}$
$= \mathop {\lim }\limits_{x \to 0} \frac{(1+x)-1}{x(\sqrt{1+x}+1)}$
$= \mathop {\lim }\limits_{x \to 0} \frac{x}{x(\sqrt{1+x}+1)}$
$= \mathop {\lim }\limits_{x \to 0} \frac{1}{\sqrt{1+x}+1}$
Substituting $x = 0$,we get:
$\frac{1}{\sqrt{1+0}+1} = \frac{1}{1+1} = \frac{1}{2}$

(A) To evaluate the limit $\mathop {\lim }\limits_{x \to 3} (x+3)$,we substitute the value $x = 3$ directly into the expression.
$\mathop {\lim }\limits_{x \to 3} (x+3) = 3 + 3 = 6$.

(A) To evaluate the limit $\mathop {\lim }\limits_{x \to \pi } \left(x-\frac{22}{7}\right)$,we substitute $x = \pi$ directly into the expression because the function is continuous at $x = \pi$.
Substituting the value,we get:
$\mathop {\lim }\limits_{x \to \pi } \left(x-\frac{22}{7}\right) = \pi - \frac{22}{7}$.

(A) To evaluate the limit $\mathop {\lim }\limits_{r \to 1} \pi r^{2}$,we substitute $r = 1$ into the expression.
$\mathop {\lim }\limits_{r \to 1} \pi r^{2} = \pi(1)^{2}$
$= \pi(1) = \pi$

(A) To evaluate the limit $\mathop {\lim }\limits_{x \to 1} \frac{4x+3}{x-2}$,we substitute $x = 1$ directly into the expression because the denominator is non-zero at $x = 1$.
$\mathop {\lim }\limits_{x \to 1} \frac{4x+3}{x-2} = \frac{4(1)+3}{1-2}$
$= \frac{4+3}{-1}$
$= \frac{7}{-1} = -7$

(B) To evaluate the limit $\mathop {\lim }\limits_{x \to -1} \frac{x^{10}+x^{5}+1}{x-1}$,we substitute $x = -1$ directly into the expression since it is a continuous function at $x = -1$.
Substituting $x = -1$:
$\frac{(-1)^{10}+(-1)^{5}+1}{-1-1}$
$= \frac{1 - 1 + 1}{-2}$
$= \frac{1}{-2}$
$= -\frac{1}{2}$

(B) We need to evaluate the limit: $\mathop {\lim }\limits_{x \to 0} \frac{(x+1)^{5}-1}{x}$
Let $y = x+1$. As $x \to 0$,$y \to 1$.
Substituting $y$ into the expression,we get:
$\mathop {\lim }\limits_{y \to 1} \frac{y^{5}-1}{y-1}$
Using the standard limit formula $\mathop {\lim }\limits_{x \to a} \frac{x^{n}-a^{n}}{x-a} = n a^{n-1}$:
Here $n = 5$ and $a = 1$.
$= 5 \times (1)^{5-1}$
$= 5 \times 1 = 5$
Therefore,$\mathop {\lim }\limits_{x \to 0} \frac{(x+1)^{5}-1}{x} = 5$

(A) At $x=2$,the expression takes the indeterminate form $\frac{0}{0}$.
Factorizing the numerator: $3x^{2}-x-10 = 3x^{2}-6x+5x-10 = 3x(x-2)+5(x-2) = (x-2)(3x+5)$.
Factorizing the denominator: $x^{2}-4 = (x-2)(x+2)$.
Thus,$\mathop {\lim }\limits_{x \to 2} \frac{(x-2)(3x+5)}{(x-2)(x+2)} = \mathop {\lim }\limits_{x \to 2} \frac{3x+5}{x+2}$.
Substituting $x=2$: $\frac{3(2)+5}{2+2} = \frac{6+5}{4} = \frac{11}{4}$.

(A) At $x=3$,the given expression takes the indeterminate form $\frac{0}{0}$.
$\mathop {\lim }\limits_{x \to 3} \frac{x^{4}-81}{2 x^{2}-5 x-3} = \mathop {\lim }\limits_{x \to 3} \frac{(x^2-9)(x^2+9)}{(2x+1)(x-3)}$
$= \mathop {\lim }\limits_{x \to 3} \frac{(x-3)(x+3)(x^2+9)}{(2x+1)(x-3)}$
$= \mathop {\lim }\limits_{x \to 3} \frac{(x+3)(x^2+9)}{2x+1}$
$= \frac{(3+3)(3^2+9)}{2(3)+1}$
$= \frac{6 \times 18}{7} = \frac{108}{7}$

(A) To evaluate the limit $\mathop {\lim }\limits_{x \to 0} \frac{ax+b}{cx+1}$,we substitute $x = 0$ directly into the expression.
$\mathop {\lim }\limits_{x \to 0} \frac{ax+b}{cx+1} = \frac{a(0)+b}{c(0)+1}$
$= \frac{0+b}{0+1}$
$= \frac{b}{1} = b$

(B) We are given the limit $\mathop {\lim }\limits_{z \to 1} \frac{z^{1/3}-1}{z^{1/6}-1}$.
At $z=1$,the expression takes the indeterminate form $\frac{0}{0}$.
Let $z^{1/6} = x$. As $z \to 1$,$x \to 1$.
Then $z^{1/3} = (z^{1/6})^2 = x^2$.
The limit becomes $\mathop {\lim }\limits_{x \to 1} \frac{x^2-1}{x-1}$.
Using the standard formula $\mathop {\lim }\limits_{x \to a} \frac{x^n-a^n}{x-a} = na^{n-1}$,where $n=2$ and $a=1$:
$\mathop {\lim }\limits_{x \to 1} \frac{x^2-1^2}{x-1} = 2(1)^{2-1} = 2(1) = 2$.
Thus,the value of the limit is $2$.

(A) To evaluate the limit $\mathop {\lim }\limits_{x \to 1} \frac{a x^{2}+b x+c}{c x^{2}+b x+a}$,we substitute $x = 1$ directly into the expression since the denominator is not zero at $x = 1$ (given $a+b+c \neq 0$).
$\mathop {\lim }\limits_{x \to 1} \frac{a x^{2}+b x+c}{c x^{2}+b x+a} = \frac{a(1)^{2}+b(1)+c}{c(1)^{2}+b(1)+a}$
$= \frac{a+b+c}{c+b+a}$
Since $a+b+c \neq 0$,we can cancel the numerator and the denominator.
$= 1$

(B) Given limit: $\mathop {\lim }\limits_{x \to -2} \frac{\frac{1}{x} + \frac{1}{2}}{x + 2}$
At $x = -2$,the expression takes the indeterminate form $\frac{0}{0}$.
Simplify the numerator:
$\frac{1}{x} + \frac{1}{2} = \frac{2 + x}{2x}$
Now substitute this back into the limit:
$\mathop {\lim }\limits_{x \to -2} \frac{\frac{2 + x}{2x}}{x + 2} = \mathop {\lim }\limits_{x \to -2} \frac{2 + x}{2x(x + 2)}$
Cancel the common factor $(x + 2)$:
$= \mathop {\lim }\limits_{x \to -2} \frac{1}{2x}$
Evaluate the limit by substituting $x = -2$:
$= \frac{1}{2(-2)} = -\frac{1}{4}$

(A) To evaluate the limit $\mathop {\lim }\limits_{x \to 0} \frac{\cos x}{\pi - x}$,we can use direct substitution since the function is continuous at $x = 0$.
Substituting $x = 0$ into the expression:
$\mathop {\lim }\limits_{x \to 0} \frac{\cos x}{\pi - x} = \frac{\cos(0)}{\pi - 0}$
Since $\cos(0) = 1$,we get:
$= \frac{1}{\pi}$

(A) Given limit: $\mathop {\lim }\limits_{x \to 0} \frac{ax + x \cos x}{b \sin x}$
At $x = 0$,the function takes the indeterminate form $\frac{0}{0}$.
We can rewrite the expression as:
$\mathop {\lim }\limits_{x \to 0} \frac{x(a + \cos x)}{b \sin x} = \frac{1}{b} \mathop {\lim }\limits_{x \to 0} \left( \frac{x}{\sin x} \right) \times \mathop {\lim }\limits_{x \to 0} (a + \cos x)$
Using the standard limit $\mathop {\lim }\limits_{x \to 0} \frac{\sin x}{x} = 1$,we have $\mathop {\lim }\limits_{x \to 0} \frac{x}{\sin x} = 1$.
Substituting the values:
$= \frac{1}{b} \times 1 \times (a + \cos 0)$
$= \frac{1}{b} \times (a + 1)$
$= \frac{a+1}{b}$

(A) We are given the limit: $\mathop {\lim }\limits_{x \to 0} x \sec x$.
Since $\sec x = \frac{1}{\cos x}$,we can rewrite the expression as:
$\mathop {\lim }\limits_{x \to 0} \frac{x}{\cos x}$.
Now,substitute $x = 0$ into the expression:
$\frac{0}{\cos 0} = \frac{0}{1} = 0$.
Thus,the value of the limit is $0$.

(A) At $x=0,$ the given function takes the indeterminate form $\infty - \infty$.
$\mathop {\lim }\limits_{x \to 0} (\text{cosec}\,x - \cot x) = \mathop {\lim }\limits_{x \to 0} \left(\frac{1}{\sin x} - \frac{\cos x}{\sin x}\right)$
$= \mathop {\lim }\limits_{x \to 0} \left(\frac{1 - \cos x}{\sin x}\right)$
Using the half-angle identities $1 - \cos x = 2\sin^2(x/2)$ and $\sin x = 2\sin(x/2)\cos(x/2)$:
$= \mathop {\lim }\limits_{x \to 0} \frac{2\sin^2(x/2)}{2\sin(x/2)\cos(x/2)}$
$= \mathop {\lim }\limits_{x \to 0} \tan(x/2)$
$= \tan(0) = 0$

(N/A) The given function is $f(x) = \begin{cases} 2x+3, & x \leq 0 \\ 3(x+1), & x > 0 \end{cases}$
For $\mathop {\lim }\limits_{x \to 0} f(x)$:
Left-hand limit: $\mathop {\lim }\limits_{x \to 0^-} f(x) = \mathop {\lim }\limits_{x \to 0} (2x+3) = 2(0)+3 = 3$
Right-hand limit: $\mathop {\lim }\limits_{x \to 0^+} f(x) = \mathop {\lim }\limits_{x \to 0} 3(x+1) = 3(0+1) = 3$
Since the left-hand limit equals the right-hand limit,$\mathop {\lim }\limits_{x \to 0} f(x) = 3$.
For $\mathop {\lim }\limits_{x \to 1} f(x)$:
Since $x=1$ is in the domain $x > 0$,we use the function $f(x) = 3(x+1)$.
$\mathop {\lim }\limits_{x \to 1} f(x) = \mathop {\lim }\limits_{x \to 1} 3(x+1) = 3(1+1) = 6$.

(D) The given function is $f(x) = \begin{cases} x^{2}-1, & x \leq 1 \\ -x-1, & x > 1 \end{cases}$
To find the limit as $x \to 1$,we evaluate the left-hand limit and the right-hand limit.
Left-hand limit:
$\mathop {\lim }\limits_{x \to 1^-} f(x) = \mathop {\lim }\limits_{x \to 1} (x^{2}-1) = 1^{2}-1 = 0$
Right-hand limit:
$\mathop {\lim }\limits_{x \to 1^+} f(x) = \mathop {\lim }\limits_{x \to 1} (-x-1) = -1-1 = -2$
Since $\mathop {\lim }\limits_{x \to 1^-} f(x) \neq \mathop {\lim }\limits_{x \to 1^+} f(x)$,
Therefore,$\mathop {\lim }\limits_{x \to 1} f(x)$ does not exist.

(D) The given function is $f(x) = \begin{cases} \frac{|x|}{x}, & x \neq 0 \\ 0, & x=0 \end{cases}$
First,we find the left-hand limit $(LHL)$:
$\mathop {\lim }\limits_{x \to 0^-} f(x) = \mathop {\lim }\limits_{x \to 0^-} \frac{|x|}{x}$
Since $x < 0$,$|x| = -x$,so:
$\mathop {\lim }\limits_{x \to 0^-} \frac{-x}{x} = \mathop {\lim }\limits_{x \to 0^-} (-1) = -1$
Next,we find the right-hand limit $(RHL)$:
$\mathop {\lim }\limits_{x \to 0^+} f(x) = \mathop {\lim }\limits_{x \to 0^+} \frac{|x|}{x}$
Since $x > 0$,$|x| = x$,so:
$\mathop {\lim }\limits_{x \to 0^+} \frac{x}{x} = \mathop {\lim }\limits_{x \to 0^+} (1) = 1$
Since the left-hand limit $(-1)$ is not equal to the right-hand limit $(1)$,the limit $\mathop {\lim }\limits_{x \to 0} f(x)$ does not exist.

(D) To find $\mathop {\lim }\limits_{x \to 0} f(x)$,we evaluate the left-hand limit $(LHL)$ and the right-hand limit $(RHL)$.
$LHL$: $\mathop {\lim }\limits_{x \to 0^-} f(x) = \mathop {\lim }\limits_{x \to 0^-} \frac{x}{|x|}$.
Since $x < 0$,$|x| = -x$,so $\mathop {\lim }\limits_{x \to 0^-} \frac{x}{-x} = \mathop {\lim }\limits_{x \to 0^-} (-1) = -1$.
$RHL$: $\mathop {\lim }\limits_{x \to 0^+} f(x) = \mathop {\lim }\limits_{x \to 0^+} \frac{x}{|x|}$.
Since $x > 0$,$|x| = x$,so $\mathop {\lim }\limits_{x \to 0^+} \frac{x}{x} = \mathop {\lim }\limits_{x \to 0^+} (1) = 1$.
Since $\mathop {\lim }\limits_{x \to 0^-} f(x) \neq \mathop {\lim }\limits_{x \to 0^+} f(x)$,the limit $\mathop {\lim }\limits_{x \to 0} f(x)$ does not exist.

(A) The given function is $f(x)=|x|-5$.
Since $x$ approaches $5$,we consider the neighborhood of $x=5$ where $x > 0$.
In this interval,$|x| = x$.
Therefore,$\mathop {\lim }\limits_{x \to 5} f(x) = \mathop {\lim }\limits_{x \to 5} (x-5)$.
Substituting $x=5$,we get $5-5 = 0$.
Thus,$\mathop {\lim }\limits_{x \to 5} f(x) = 0$.

The given function is $f(x) = (x - a_{1})(x - a_{2}) \dots (x - a_{n})$.
$\lim_{x \to a_{1}} f(x) = \lim_{x \to a_{1}} [(x - a_{1})(x - a_{2}) \dots (x - a_{n})]$
$= (a_{1} - a_{1})(a_{1} - a_{2}) \dots (a_{1} - a_{n}) = 0 \times (a_{1} - a_{2}) \dots (a_{1} - a_{n}) = 0$.
Therefore,$\lim_{x \to a_{1}} f(x) = 0$.
Now,for $a \neq a_{1}, a_{2}, \dots, a_{n}$,
$\lim_{x \to a} f(x) = \lim_{x \to a} [(x - a_{1})(x - a_{2}) \dots (x - a_{n})]$
$= (a - a_{1})(a - a_{2}) \dots (a - a_{n})$.
Therefore,$\lim_{x \to a} f(x) = (a - a_{1})(a - a_{2}) \dots (a - a_{n})$.

(A) The given function is $f(x) = \begin{cases} |x|+1, & x < 0 \\ 0, & x = 0 \\ |x|-1, & x > 0 \end{cases}$.
Case $1$: $a = 0$.
$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-x+1) = 1$.
$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x-1) = -1$.
Since $1 \neq -1$,the limit does not exist at $a = 0$.
Case $2$: $a < 0$.
$\lim_{x \to a^-} f(x) = \lim_{x \to a^-} (-x+1) = -a+1$.
$\lim_{x \to a^+} f(x) = \lim_{x \to a^+} (-x+1) = -a+1$.
Since the left-hand limit equals the right-hand limit,the limit exists for all $a < 0$.
Case $3$: $a > 0$.
$\lim_{x \to a^-} f(x) = \lim_{x \to a^-} (x-1) = a-1$.
$\lim_{x \to a^+} f(x) = \lim_{x \to a^+} (x-1) = a-1$.
Since the left-hand limit equals the right-hand limit,the limit exists for all $a > 0$.
Conclusion: The limit exists for all $a \neq 0$.

(D) The given limit is of the form $1^{\infty}$.
We use the formula $\lim \limits_{x \rightarrow a} f(x)^{g(x)} = e^{\lim \limits_{x \rightarrow a} g(x)(f(x)-1)}$.
Here,$f(x) = \tan(\frac{\pi}{4} + x)$ and $g(x) = \frac{1}{x}$.
$\lim \limits_{x \rightarrow 0} \left(\tan \left(\frac{\pi}{4}+x\right)\right)^{\frac{1}{x}} = e^{\lim \limits_{x \rightarrow 0} \frac{1}{x} \left(\tan \left(\frac{\pi}{4}+x\right)-1\right)}$.
Using the identity $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,we have $\tan(\frac{\pi}{4}+x) = \frac{1+\tan x}{1-\tan x}$.
So,$\tan(\frac{\pi}{4}+x)-1 = \frac{1+\tan x - (1-\tan x)}{1-\tan x} = \frac{2\tan x}{1-\tan x}$.
Thus,the exponent becomes $\lim \limits_{x \rightarrow 0} \frac{2\tan x}{x(1-\tan x)} = \lim \limits_{x \rightarrow 0} 2 \cdot \frac{\tan x}{x} \cdot \frac{1}{1-\tan x} = 2 \cdot 1 \cdot 1 = 2$.
Therefore,the limit is $e^{2}$.

(B) For the limit to exist,the left-hand limit $(LHL)$ must equal the right-hand limit $(RHL)$.
$LHL = \lim_{x \rightarrow 0^{-}} \left| \frac{1-x+(-x)}{\lambda-x+(-1)} \right| = \left| \frac{1}{\lambda-1} \right|$
$RHL = \lim_{x \rightarrow 0^{+}} \left| \frac{1-x+x}{\lambda-x+0} \right| = \left| \frac{1}{\lambda} \right|$
Equating $LHL$ and $RHL$:
$\left| \frac{1}{\lambda-1} \right| = \left| \frac{1}{\lambda} \right| \Rightarrow |\lambda| = |\lambda-1|$
Squaring both sides: $\lambda^2 = \lambda^2 - 2\lambda + 1$ $\Rightarrow 2\lambda = 1$ $\Rightarrow \lambda = \frac{1}{2}$.
Substituting $\lambda = \frac{1}{2}$ into the expression for $L$:
$L = \left| \frac{1}{1/2} \right| = 2$.

(B) The given expression is $\lim_{x}$ ${\rightarrow 0} \frac{1}{x^{8}} \left( 1 - \cos \frac{x^{2}}{2} - \cos \frac{x^{2}}{4} + \cos \frac{x^{2}}{2} \cos \frac{x^{2}}{4} \right)$.
Factoring the numerator,we get $\lim_{x \rightarrow 0} \frac{(1 - \cos \frac{x^{2}}{2})(1 - \cos \frac{x^{2}}{4})}{x^{8}}$.
Using the limit formula $\lim_{\theta \rightarrow 0} \frac{1 - \cos \theta}{\theta^{2}} = \frac{1}{2}$,we rewrite the expression as:
$\lim_{x}$ ${\rightarrow 0} \left( \frac{1 - \cos \frac{x^{2}}{2}}{(x^{2}/2)^{2} \cdot 4} \right) \cdot \left( \frac{1 - \cos \frac{x^{2}}{4}}{(x^{2}/4)^{2} \cdot 16} \right) = \frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{2} \cdot \frac{1}{16} = \frac{1}{256}$.
Since $\frac{1}{256} = \frac{1}{2^{8}} = 2^{-8}$,we have $2^{-8} = 2^{-k}$.
Therefore,$k = 8$.

(D) For $x \in (0, 1)$,the greatest integer function $[x] = 0$.
Substituting this into the expression,we get:
$\lim _{x \rightarrow 0^{+}} \frac{\cos ^{-1}(x-0) \cdot \sin ^{-1}(x-0)}{x(1-x^2)}$
$= \lim _{x \rightarrow 0^{+}} \left( \frac{\cos ^{-1} x}{1-x^2} \cdot \frac{\sin ^{-1} x}{x} \right)$
As $x \rightarrow 0^{+}$,$\cos ^{-1} x \rightarrow \frac{\pi}{2}$,$1-x^2 \rightarrow 1$,and $\frac{\sin ^{-1} x}{x} \rightarrow 1$.
Therefore,the limit is $\frac{\pi}{2} \cdot 1 \cdot 1 = \frac{\pi}{2}$.

(D) The given limit is of the form $1^{\infty}$.
Let $L = \lim _{n \rightarrow \infty}\left(1+\frac{H_n}{n^{2}}\right)^{n}$,where $H_n = 1+\frac{1}{2}+\ldots+\frac{1}{n}$.
Using the formula $\lim _{n \rightarrow \infty}(1+f(n))^{g(n)} = e^{\lim _{n \rightarrow \infty} f(n)g(n)}$,we get:
$L = \exp \left(\lim _{n \rightarrow \infty} \frac{H_n}{n^2} \cdot n\right) = \exp \left(\lim _{n \rightarrow \infty} \frac{H_n}{n}\right)$.
We know that $H_n \approx \ln(n) + \gamma$ (where $\gamma$ is the Euler-Mascheroni constant).
Thus,$\lim _{n \rightarrow \infty} \frac{\ln(n) + \gamma}{n} = 0$.
Therefore,$L = e^0 = 1$.

(A) Let $L = \lim_{h \rightarrow 0} 2 \left\{ \frac{\sqrt{3} \sin (\frac{\pi}{6} + h) - \cos (\frac{\pi}{6} + h)}{\sqrt{3} h (\sqrt{3} \cos h - \sin h)} \right\}$.
Using the expansion $\sin(A+B) = \sin A \cos B + \cos A \sin B$ and $\cos(A+B) = \cos A \cos B - \sin A \sin B$:
Numerator $= \sqrt{3} (\sin \frac{\pi}{6} \cos h + \cos \frac{\pi}{6} \sin h) - (\cos \frac{\pi}{6} \cos h - \sin \frac{\pi}{6} \sin h)$
$= \sqrt{3} (\frac{1}{2} \cos h + \frac{\sqrt{3}}{2} \sin h) - (\frac{\sqrt{3}}{2} \cos h - \frac{1}{2} \sin h)$
$= \frac{\sqrt{3}}{2} \cos h + \frac{3}{2} \sin h - \frac{\sqrt{3}}{2} \cos h + \frac{1}{2} \sin h = 2 \sin h$.
Substituting back into the limit:
$L = \lim_{h \rightarrow 0} 2 \left( \frac{2 \sin h}{\sqrt{3} h (\sqrt{3} \cos h - \sin h)} \right) = \frac{4}{\sqrt{3}} \lim_{h \rightarrow 0} \left( \frac{\sin h}{h} \right) \cdot \lim_{h \rightarrow 0} \left( \frac{1}{\sqrt{3} \cos h - \sin h} \right)$.
Since $\lim_{h \rightarrow 0} \frac{\sin h}{h} = 1$ and $\lim_{h \rightarrow 0} (\sqrt{3} \cos h - \sin h) = \sqrt{3}(1) - 0 = \sqrt{3}$:
$L = \frac{4}{\sqrt{3}} \cdot 1 \cdot \frac{1}{\sqrt{3}} = \frac{4}{3}$.

(A) Let $S = \lim _{x \rightarrow 2} \sum_{n=1}^{9} \frac{x}{n(n+1) x^{2}+2(2 n+1) x+4}$.
Substitute $x = 2$ into the expression:
$S = \sum_{n=1}^{9} \frac{2}{n(n+1)(4) + 2(2n+1)(2) + 4}$
$S = \sum_{n=1}^{9} \frac{2}{4n^2 + 4n + 8n + 4 + 4} = \sum_{n=1}^{9} \frac{2}{4n^2 + 12n + 8}$
$S = \sum_{n=1}^{9} \frac{2}{4(n^2 + 3n + 2)} = \sum_{n=1}^{9} \frac{1}{2(n+1)(n+2)}$
Using partial fractions:
$S = \frac{1}{2} \sum_{n=1}^{9} \left( \frac{1}{n+1} - \frac{1}{n+2} \right)$
This is a telescoping series:
$S = \frac{1}{2} \left[ \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \dots + \left( \frac{1}{10} - \frac{1}{11} \right) \right]$
$S = \frac{1}{2} \left( \frac{1}{2} - \frac{1}{11} \right) = \frac{1}{2} \left( \frac{11-2}{22} \right) = \frac{1}{2} \left( \frac{9}{22} \right) = \frac{9}{44}$.

(B) Given $\lim _{x \rightarrow \infty}\left(\sqrt{x^{2}-x+1}-a x\right)=b$.
For the limit to exist as a finite value,the degree of the expression must be balanced,so $a$ must be $1$.
Rationalizing the expression:
$\lim _{x \rightarrow \infty} \frac{(\sqrt{x^{2}-x+1}-ax)(\sqrt{x^{2}-x+1}+ax)}{\sqrt{x^{2}-x+1}+ax} = b$
$\lim _{x \rightarrow \infty} \frac{x^{2}-x+1-a^{2}x^{2}}{\sqrt{x^{2}-x+1}+ax} = b$
$\lim _{x \rightarrow \infty} \frac{(1-a^{2})x^{2}-x+1}{\sqrt{x^{2}-x+1}+ax} = b$
Since the limit is finite,the coefficient of $x^{2}$ must be $0$,so $1-a^{2}=0$. Since $x \rightarrow \infty$,we take $a=1$.
Now,$\lim _{x \rightarrow \infty} \frac{-x+1}{\sqrt{x^{2}-x+1}+x} = b$
Dividing numerator and denominator by $x$:
$\lim _{x}$ ${\rightarrow \infty} \frac{-1+\frac{1}{x}}{\sqrt{1-\frac{1}{x}+\frac{1}{x^{2}}}+1} = \frac{-1}{1+1} = -\frac{1}{2}$
Thus,$b = -\frac{1}{2}$.
The ordered pair $(a, b)$ is $\left(1, -\frac{1}{2}\right)$.

(C) Let $L = \lim _{x \rightarrow 0}(2-\cos x \sqrt{\cos 2 x})^{\frac{x+2}{x^{2}}}$.
This is of the form $1^{\infty}$.
Using the formula $\lim _{x \rightarrow 0} f(x)^{g(x)} = e^{\lim _{x \rightarrow 0} (f(x)-1)g(x)}$,we get:
$L = e^{\lim _{x \rightarrow 0} (2-\cos x \sqrt{\cos 2 x}-1) \left(\frac{x+2}{x^{2}}\right)}$
$L = e^{\lim _{x \rightarrow 0} (1-\cos x \sqrt{\cos 2 x}) \left(\frac{x+2}{x^{2}}\right)}$
As $x \rightarrow 0$,$\frac{x+2}{x^2} \approx \frac{2}{x^2}$.
So,$L = e^{\lim _{x \rightarrow 0} \frac{1-\cos x \sqrt{\cos 2 x}}{x^2} \times 2}$.
Let $f(x) = 1-\cos x \sqrt{\cos 2 x}$. Using Taylor series expansion:
$\cos x \approx 1 - \frac{x^2}{2}$ and $\sqrt{\cos 2 x} = (1 - 2x^2)^{1/2} \approx 1 - x^2$.
$f(x) \approx 1 - (1 - \frac{x^2}{2})(1 - x^2) = 1 - (1 - x^2 - \frac{x^2}{2} + \frac{x^4}{2}) \approx \frac{3x^2}{2}$.
Thus,$\lim _{x \rightarrow 0} \frac{f(x)}{x^2} = \frac{3}{2}$.
Therefore,$L = e^{\frac{3}{2} \times 2} = e^3$.
Comparing with $e^a$,we get $a = 3$.

(A) Let $L = \lim_{x \rightarrow \frac{\pi}{2}} \tan^{2} x (\sqrt{2 \sin^{2} x + 3 \sin x + 4} - \sqrt{\sin^{2} x + 6 \sin x + 2})$.
Rationalizing the expression inside the limit:
$L = \lim_{x \rightarrow \frac{\pi}{2}} \tan^{2} x \cdot \frac{(2 \sin^{2} x + 3 \sin x + 4) - (\sin^{2} x + 6 \sin x + 2)}{\sqrt{2 \sin^{2} x + 3 \sin x + 4} + \sqrt{\sin^{2} x + 6 \sin x + 2}}$
$L = \lim_{x \rightarrow \frac{\pi}{2}} \tan^{2} x \cdot \frac{\sin^{2} x - 3 \sin x + 2}{\sqrt{2 \sin^{2} x + 3 \sin x + 4} + \sqrt{\sin^{2} x + 6 \sin x + 2}}$
As $x \rightarrow \frac{\pi}{2}$,$\sin x \rightarrow 1$. The denominator approaches $\sqrt{2+3+4} + \sqrt{1+6+2} = \sqrt{9} + \sqrt{9} = 6$.
$L = \frac{1}{6} \lim_{x \rightarrow \frac{\pi}{2}} \tan^{2} x (\sin x - 1)(\sin x - 2)$
Since $\sin x - 2 \rightarrow -1$ as $x \rightarrow \frac{\pi}{2}$,we have:
$L = \frac{1}{6} \lim_{x \rightarrow \frac{\pi}{2}} \frac{\sin^{2} x}{\cos^{2} x} (\sin x - 1)(-1) = \frac{1}{6} \lim_{x \rightarrow \frac{\pi}{2}} \frac{\sin^{2} x (1 - \sin x)}{1 - \sin^{2} x}$
$L = \frac{1}{6} \lim_{x \rightarrow \frac{\pi}{2}} \frac{\sin^{2} x (1 - \sin x)}{(1 - \sin x)(1 + \sin x)} = \frac{1}{6} \cdot \frac{1^2}{1+1} = \frac{1}{12}$.

(C) We evaluate the limit: $\lim\limits _{x \rightarrow 0} \frac{\cos (\sin x)-\cos x}{x^{4}}$.
Using the Taylor series expansion for $\cos \theta = 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \dots$:
$\sin x = x - \frac{x^3}{6} + O(x^5)$
$\cos(\sin x) = 1 - \frac{(x - \frac{x^3}{6})^2}{2} + \frac{(x)^4}{24} + O(x^6) = 1 - \frac{x^2 - \frac{x^4}{3}}{2} + \frac{x^4}{24} + O(x^6) = 1 - \frac{x^2}{2} + \frac{x^4}{6} + \frac{x^4}{24} + O(x^6) = 1 - \frac{x^2}{2} + \frac{5x^4}{24} + O(x^6)$.
$\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} + O(x^6)$.
Subtracting these:
$\cos(\sin x) - \cos x = (1 - \frac{x^2}{2} + \frac{5x^4}{24}) - (1 - \frac{x^2}{2} + \frac{x^4}{24}) + O(x^6) = \frac{4x^4}{24} = \frac{x^4}{6}$.
Therefore,$\lim\limits _{x \rightarrow 0} \frac{\frac{x^4}{6}}{x^4} = \frac{1}{6}$.

(D) Let $f(x) = \frac{\sin(\cos^{-1} x) - x}{1 - \tan(\cos^{-1} x)}$.
We know that $\cos^{-1} x = \sin^{-1}(\sqrt{1-x^2})$ and $\cos^{-1} x = \tan^{-1}(\frac{\sqrt{1-x^2}}{x})$.
Substituting these into the expression:
$\lim \limits_{x}$ ${\rightarrow \frac{1}{\sqrt{2}}} \frac{\sin(\sin^{-1}(\sqrt{1-x^2})) - x}{1 - \tan(\tan^{-1}(\frac{\sqrt{1-x^2}}{x}))}$
$= \lim \limits_{x \rightarrow \frac{1}{\sqrt{2}}} \frac{\sqrt{1-x^2} - x}{1 - \frac{\sqrt{1-x^2}}{x}}$
$= \lim \limits_{x \rightarrow \frac{1}{\sqrt{2}}} \frac{\sqrt{1-x^2} - x}{\frac{x - \sqrt{1-x^2}}{x}}$
$= \lim \limits_{x \rightarrow \frac{1}{\sqrt{2}}} \frac{- (x - \sqrt{1-x^2})}{\frac{x - \sqrt{1-x^2}}{x}}$
$= \lim \limits_{x \rightarrow \frac{1}{\sqrt{2}}} (-x) = -\frac{1}{\sqrt{2}}$.

(C) Let $T_r = \tan^{-1}\left(\frac{1}{r^2+3r+3}\right)$.
We can rewrite the argument as $\frac{(r+2)-(r+1)}{1+(r+2)(r+1)}$.
Thus,$T_r = \tan^{-1}(r+2) - \tan^{-1}(r+1)$.
The sum $S_n = \sum_{r=1}^{n} T_r$ is a telescoping sum:
$S_n = (\tan^{-1}3 - \tan^{-1}2) + (\tan^{-1}4 - \tan^{-1}3) + \dots + (\tan^{-1}(n+2) - \tan^{-1}(n+1))$.
$S_n = \tan^{-1}(n+2) - \tan^{-1}2$.
Using the formula $\tan^{-1}x - \tan^{-1}y = \tan^{-1}\left(\frac{x-y}{1+xy}\right)$:
$S_n = \tan^{-1}\left(\frac{(n+2)-2}{1+(n+2)(2)}\right) = \tan^{-1}\left(\frac{n}{2n+5}\right)$.
Now,we evaluate the limit:
$\lim\limits_{n \rightarrow \infty} 6 \tan(S_n) = \lim\limits_{n \rightarrow \infty} 6 \tan\left(\tan^{-1}\left(\frac{n}{2n+5}\right)\right)$.
$= \lim\limits_{n \rightarrow \infty} \frac{6n}{2n+5} = \frac{6}{2} = 3$.

(D) Let $L = \lim_{x \rightarrow 1} \frac{(x^{2}-1) \sin^{2}(\pi x)}{x^{4}-2x^{3}+2x-1}$.
First,factor the denominator: $x^{4}-2x^{3}+2x-1 = (x^{4}-1) - 2x(x^{2}-1) = (x^{2}-1)(x^{2}+1) - 2x(x^{2}-1) = (x^{2}-1)(x^{2}-2x+1) = (x^{2}-1)(x-1)^{2}$.
Substitute this into the limit: $L = \lim_{x \rightarrow 1} \frac{(x^{2}-1) \sin^{2}(\pi x)}{(x^{2}-1)(x-1)^{2}} = \lim_{x \rightarrow 1} \frac{\sin^{2}(\pi x)}{(x-1)^{2}}$.
Using the property $\sin(\pi x) = \sin(\pi - \pi x) = \sin(\pi(1-x))$,we get:
$L = \lim_{x \rightarrow 1} \left( \frac{\sin(\pi(1-x))}{\pi(1-x)} \cdot \pi \right)^{2} = \pi^{2} \cdot \left( \lim_{x \rightarrow 1} \frac{\sin(\pi(1-x))}{\pi(1-x)} \right)^{2} = \pi^{2} \cdot (1)^{2} = \pi^{2}$.

(B) Let $LHS = \lim_{n \rightarrow \infty} \frac{(n+1)^{k-1}}{n^{k+1}} [nk \cdot n + \frac{n(n+1)}{2}]$
$= \lim_{n \rightarrow \infty} \frac{(n+1)^{k-1}}{n^{k+1}} \cdot n^2 [k + \frac{1 + \frac{1}{n}}{2}]$
$= \lim_{n \rightarrow \infty} (1 + \frac{1}{n})^{k-1} \cdot (k + \frac{1 + \frac{1}{n}}{2}) = k + \frac{1}{2}$.
Let $RHS = \lim_{n \rightarrow \infty} \frac{1}{n^{k+1}} \sum_{i=1}^{n} i^k = \int_{0}^{1} x^k dx = \frac{1}{k+1}$.
Given $LHS = 33 \cdot RHS$,so $k + \frac{1}{2} = 33 \cdot \frac{1}{k+1}$.
$(2k + 1)(k + 1) = 66$.
$2k^2 + 3k + 1 = 66 \implies 2k^2 + 3k - 65 = 0$.
$(2k + 13)(k - 5) = 0$.
Since $k$ is an integral value,$k = 5$.

(B) Let $f(x) = \frac{(x+2 \cos x)^{3}+2(x+2 \cos x)^{2}+3 \sin (x+2 \cos x)}{(x+2)^{3}+2(x+2)^{2}+3 \sin (x+2)}$.
As $x \rightarrow 0$,$f(x) \rightarrow \frac{2^3 + 2(2^2) + 3 \sin 2}{2^3 + 2(2^2) + 3 \sin 2} = 1$.
This is a $1^{\infty}$ form.
Using the formula $\lim_{x \rightarrow a} f(x)^{g(x)} = e^{\lim_{x \rightarrow a} g(x)(f(x)-1)}$,we get:
$L = e^{\lim_{x \rightarrow 0} \frac{100}{x} \left( \frac{(x+2 \cos x)^{3}+2(x+2 \cos x)^{2}+3 \sin (x+2 \cos x) - (x+2)^{3}-2(x+2)^{2}-3 \sin (x+2)}{(x+2)^{3}+2(x+2)^{2}+3 \sin (x+2)} \right)}$.
Let $h(u) = u^3 + 2u^2 + 3 \sin u$. Then $f(x) = \frac{h(x+2 \cos x)}{h(x+2)}$.
As $x \rightarrow 0$,$f(x) - 1 = \frac{h(x+2 \cos x) - h(x+2)}{h(x+2)}$.
Using the derivative $h'(u) = 3u^2 + 4u + 3 \cos u$,the limit becomes $e^{100 \cdot \frac{h'(2) \cdot \lim_{x \rightarrow 0} \frac{(x+2 \cos x) - (x+2)}{x}}{h(2)}}$.
Since $\lim_{x \rightarrow 0} \frac{2 \cos x - 2}{x} = \lim_{x \rightarrow 0} \frac{-2 \sin x}{1} = 0$,the exponent is $0$.
Thus,$L = e^0 = 1$.

(A) For statement $I$: $\lim _{n \rightarrow \infty} \frac{2^n+(-2)^n}{2^n} = \lim _{n \rightarrow \infty} (1 + (-1)^n)$.
As $n \rightarrow \infty$,the expression oscillates between $1+1=2$ (for even $n$) and $1-1=0$ (for odd $n$). Therefore,the limit does not exist. Statement $I$ is true.
For statement $II$: $\lim _{n \rightarrow \infty} \frac{3^n+(-3)^n}{4^n} = \lim _{n \rightarrow \infty} \left(\frac{3}{4}\right)^n + \left(-\frac{3}{4}\right)^n$.
Since $|\frac{3}{4}| < 1$ and $|-\frac{3}{4}| < 1$,both terms approach $0$ as $n \rightarrow \infty$. Thus,the limit is $0+0=0$. Statement $II$ is false.