Concept of limits, Evaluation of algebric limits Questions in English

(B) Given $f(x) = \lim_{n \rightarrow \infty} n(x^{1/n} - 1)$.
Let $h = \frac{1}{n}$. As $n \rightarrow \infty$,$h \rightarrow 0$.
Then $f(x) = \lim_{h \rightarrow 0} \frac{x^h - 1}{h}$.
This is the standard limit definition of the derivative of $a^x$ at $x=0$,which is $\ln(x)$.
Thus,$f(x) = \ln(x)$.
Now,$f(xy) = \ln(xy) = \ln(x) + \ln(y) = f(x) + f(y)$.

(C) We have,$\lim _{x \rightarrow 1}\left(\frac{1+x}{2+x}\right)^{\frac{1-\sqrt{x}}{1-x}}$
$= \lim _{x \rightarrow 1}\left(\frac{1+x}{2+x}\right)^{\frac{1-\sqrt{x}}{(1+\sqrt{x})(1-\sqrt{x})}}$
$= \lim _{x \rightarrow 1}\left(\frac{1+x}{2+x}\right)^{\frac{1}{1+\sqrt{x}}}$
$= \left(\frac{1+1}{2+1}\right)^{\frac{1}{1+1}} = \left(\frac{2}{3}\right)^{\frac{1}{2}} = \sqrt{\frac{2}{3}}$

(D) Let $x = 1 + h$. As $x \rightarrow 1$,$h \rightarrow 0$.
Substituting this into the limit:
$\lim_{h \rightarrow 0} \frac{\sin(e^{(1+h)-1}-1)}{\log(1+h)} = \lim_{h \rightarrow 0} \frac{\sin(e^h-1)}{\log(1+h)}$
We know that $\lim_{u \rightarrow 0} \frac{\sin u}{u} = 1$,$\lim_{h \rightarrow 0} \frac{\log(1+h)}{h} = 1$,and $\lim_{h \rightarrow 0} \frac{e^h-1}{h} = 1$.
Rewriting the expression:
$\lim_{h \rightarrow 0} \left( \frac{\sin(e^h-1)}{e^h-1} \cdot \frac{e^h-1}{h} \cdot \frac{h}{\log(1+h)} \right)$
$= 1 \cdot 1 \cdot \frac{1}{1} = 1$.

(C) We evaluate the limit of the form $1^{\infty}$ using the formula $\lim _{x \rightarrow \infty} f(x)^{g(x)} = e^{\lim _{x \rightarrow \infty} g(x)(f(x)-1)}$.
Given $f(x) = \frac{3x-1}{3x+1}$ and $g(x) = 4x$.
$L = \lim _{x \rightarrow \infty} 4x \left( \frac{3x-1}{3x+1} - 1 \right)$
$L = \lim _{x \rightarrow \infty} 4x \left( \frac{3x-1 - (3x+1)}{3x+1} \right)$
$L = \lim _{x \rightarrow \infty} 4x \left( \frac{-2}{3x+1} \right)$
$L = \lim _{x \rightarrow \infty} \frac{-8x}{3x+1} = \lim _{x \rightarrow \infty} \frac{-8}{3 + 1/x} = -\frac{8}{3}$.
Therefore,the limit is $e^{-8/3}$.

(A) We use the standard limit formula $\operatorname{Lt}_{x \rightarrow a} [f(x)]^{g(x)} = e^{\operatorname{Lt}_{x \rightarrow a} g(x)[f(x)-1]}$ for the indeterminate form $1^{\infty}$.
Here,$f(x) = \frac{1+5x^2}{1+3x^2}$ and $g(x) = \frac{1}{x^2}$.
As $x \rightarrow 0$,$f(x) \rightarrow 1$ and $g(x) \rightarrow \infty$.
Thus,the limit is $e^{\operatorname{Lt}_{x \rightarrow 0} \frac{1}{x^2} \left( \frac{1+5x^2}{1+3x^2} - 1 \right)}$.
$= e^{\operatorname{Lt}_{x \rightarrow 0} \frac{1}{x^2} \left( \frac{1+5x^2 - (1+3x^2)}{1+3x^2} \right)}$.
$= e^{\operatorname{Lt}_{x \rightarrow 0} \frac{1}{x^2} \left( \frac{2x^2}{1+3x^2} \right)}$.
$= e^{\operatorname{Lt}_{x \rightarrow 0} \frac{2}{1+3x^2}}$.
$= e^{\frac{2}{1+0}} = e^2$.

(C) Let $L = \lim_{x \to 0} \frac{\ln(\sec(ex)) + \ln(\sec(e^{2}x)) + ... + \ln(\sec(e^{10}x))}{e^{2} - e^{2\cos x}}$.
Using the expansion $\ln(\sec \theta) \approx \frac{\theta^{2}}{2}$ as $\theta \to 0$,the numerator becomes $\sum_{k=1}^{10} \frac{(e^{k}x)^{2}}{2} = \frac{x^{2}}{2} \sum_{k=1}^{10} e^{2k}$.
The denominator is $e^{2} - e^{2\cos x} = e^{2}(1 - e^{2\cos x - 2}) \approx e^{2}(-(2\cos x - 2)) = 2e^{2}(1 - \cos x) \approx 2e^{2}(\frac{x^{2}}{2}) = e^{2}x^{2}$.
Thus,$L = \lim_{x \to 0} \frac{\frac{x^{2}}{2} \sum_{k=1}^{10} e^{2k}}{e^{2}x^{2}} = \frac{1}{2e^{2}} \sum_{k=1}^{10} (e^{2})^{k}$.
Using the geometric series sum formula $\sum_{k=1}^{n} r^{k} = \frac{r(r^{n}-1)}{r-1}$,where $r = e^{2}$ and $n = 10$:
$L = \frac{1}{2e^{2}} \cdot \frac{e^{2}((e^{2})^{10} - 1)}{e^{2} - 1} = \frac{e^{20} - 1}{2(e^{2} - 1)}$.

(B) We use the Taylor series expansion for $\sin x$ near $x = 0$: $\sin x = x - \frac{x^3}{6} + O(x^5)$.
Then,$\sin^2 x = (x - \frac{x^3}{6} + O(x^5))^2 = x^2 - 2(x)(\frac{x^3}{6}) + O(x^6) = x^2 - \frac{x^4}{3} + O(x^6)$.
Substituting this into the denominator: $x^2 - \sin^2 x = x^2 - (x^2 - \frac{x^4}{3} + O(x^6)) = \frac{x^4}{3} + O(x^6)$.
Now,substitute this into the limit expression: $\lim_{x \to 0} \frac{x^2 \sin^2 x}{x^2 - \sin^2 x} = \lim_{x \to 0} \frac{x^2 (x^2 + O(x^4))}{\frac{x^4}{3} + O(x^6)}$.
$= \lim_{x \to 0} \frac{x^4 + O(x^6)}{\frac{x^4}{3} + O(x^6)} = \frac{1}{1/3} = 3$.

(C) To evaluate the limit $f(x) = \lim_{y \to 0} \frac{(1 - \cos(xy))\tan(xy)}{y^3}$,we multiply and divide by $(xy)^3$:
$f(x) = \lim_{y \to 0} \left( \frac{1 - \cos(xy)}{(xy)^2} \cdot \frac{\tan(xy)}{xy} \cdot \frac{x^3 y^3}{y^3} \right)$
Using standard limits $\lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta^2} = \frac{1}{2}$ and $\lim_{\theta \to 0} \frac{\tan \theta}{\theta} = 1$,we get:
$f(x) = \frac{1}{2} \cdot 1 \cdot x^3 = \frac{x^3}{2}$.
Now,we need to find the number of solutions for the equation $f(x) = \sin x$,which is $\frac{x^3}{2} = \sin x$,or $x^3 = 2 \sin x$.
Let $g(x) = x^3 - 2 \sin x$. We look for roots where $g(x) = 0$.
At $x = 0$,$g(0) = 0 - 2(0) = 0$. So,$x = 0$ is a solution.
For $x > 0$,$x^3 = 2 \sin x$ has one positive solution because $x^3$ is strictly increasing and $2 \sin x$ is bounded by $2$.
For $x < 0$,let $x = -t$ where $t > 0$. Then $(-t)^3 = 2 \sin(-t) \implies -t^3 = -2 \sin t \implies t^3 = 2 \sin t$. This gives one negative solution.
Thus,there are $3$ solutions in total: $x = 0$,$x \approx 1.41$,and $x \approx -1.41$.