Chapter 2

\(\operatorname{cosec} \frac{\pi x}{2} \rightarrow 1^{+}\)when \(x \rightarrow 1\)

\(\lim _{x \rightarrow 0} \frac{a^{\sqrt{x}}-a^{1 / \sqrt{x}}}{a^{\sqrt{x}}+a^{1 / \sqrt{x}}}, a>1\)

\(\lim _{x \rightarrow \infty} \frac{\sin ^{4} x-\sin ^{2} x+1}{\cos ^{4} x-\cos ^{2} x+1}=\lim _{x \rightarrow \infty} \frac{\left(1-\cos ^{2} x\right)^{2}-\left(1-\cos ^{2} x\right)+1}{\cos ^{4} x-\cos ^{2} x+1}\)

We have $$ f(x)+g(x)+h(x)=\frac{x^{2}-4 x+17-4 x-2}{x^{2}+x-12} $$

We have $$ \lim _{x \rightarrow \pi} \frac{1+\cos ^{3} x}{\sin ^{2} x}=\lim _{x \rightarrow \pi} \frac{(1+\cos x)\left(1-\cos x+\cos ^{2} x\right)}{(1-\cos x)(1+\cos x)} $$

\(\lim _{x \rightarrow 2} \frac{\sqrt{1+\sqrt{2+x}}-\sqrt{3}}{x-2}\)

\(\lim _{x \rightarrow 2} \frac{2^{x}+2^{3-x}-6}{\sqrt{2^{-x}}-2^{1-x}}\)

\(\lim _{x \rightarrow 2}\left[\left(\frac{x(x-2)(x+2)}{(x-2)\left(x^{2}+2 x+4\right)}\right)^{-1}-\left(\frac{\sqrt{x}(\sqrt{x}+\sqrt{2})}{(\sqrt{x}-\sqrt{2})(\sqrt{x}+\sqrt{2})}-\frac{\sqrt{2}}{\sqrt{x}-\sqrt{2}}\right)^{-1}\right]\)

\(\lim _{x \rightarrow 1} \frac{(1-x)\left(1-x^{2}\right) \cdots\left(1-x^{2 n}\right)}{\left\\{(1-x)\left(1-x^{2}\right) \cdots\left(1-x^{n}\right)\right\\}^{2}}\)

\(\lim _{x \rightarrow \infty}\left(\frac{x^{3}}{3 x^{2}-4}-\frac{x^{2}}{3 x+2}\right)=\lim _{x \rightarrow \infty} \frac{x^{3}(3 x+2)-x^{2}\left(3 x^{2}-4\right)}{\left(3 x^{2}-4\right)(3 x+2)}\)

$$ \lim _{n \rightarrow \infty} \frac{n(2 n+1)^{2}}{(n+2)\left(n^{2}+3 n-1\right)}=\lim _{n \rightarrow \infty} \frac{\left(2+\frac{1}{n}\right)^{2}}{\left(1+\frac{2}{n}\right)\left(1+\frac{3}{n}-\frac{1}{n^{2}}\right)} $$

$$ \lim _{x \rightarrow \infty} \frac{(2 x+1)^{40}(4 x-1)^{5}}{(2 x+3)^{45}}=\lim _{x \rightarrow \infty} \frac{\left(2+\frac{1}{x}\right)^{40}\left(4-\frac{1}{x}\right)^{5}}{\left(2+\frac{3}{x}\right)^{45}} $$

$$ \lim _{x \rightarrow \infty}[\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}]=\lim _{x \rightarrow \infty} \frac{x+\sqrt{x+\sqrt{x}}-x}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}} $$

The given limit is $$ \lim _{x \rightarrow \infty} \frac{\frac{2}{x}+2+\frac{\sin 2 x}{x}}{\left(2+\frac{\sin 2 x}{x}\right) e^{\sin x}} $$

$$ \lim _{x \rightarrow \infty} \frac{(x+1)^{10}+(x+2)^{10}+\cdots+(x+100)^{10}}{x^{10}+10^{10}} $$

$$ \lim _{n \rightarrow \infty} \frac{n \cdot 3^{n}}{n(x-2)^{n}+n \cdot 3^{n+1}-3^{n}}=\frac{1}{3} $$

$$ \lim _{n \rightarrow \infty} n^{2}\left(x^{1 / n}-x^{\frac{1}{n+1}}\right)=\lim _{n \rightarrow \infty} n^{2} \cdot x^{\frac{1}{n+1}}\left(x^{\frac{1}{n}-\frac{1}{n+1}}-1\right) $$

Given \(g(x)=\lim _{n \rightarrow \infty} \frac{1}{\left(\frac{3}{\pi} \tan ^{-1} 2 x\right)^{2 n}+5}=0\)

\(L=\lim _{x \rightarrow \infty} \frac{\ln \left(x^{2}+e^{x}\right)}{\ln \left(x^{4}+e^{2 x}\right)}=\lim _{x \rightarrow \infty} \frac{\ln e^{x}\left(1+\frac{x^{2}}{e^{x}}\right)}{\ln e^{2 x}\left(1+\frac{x^{4}}{e^{2 x}}\right)}\)

\(\lim _{n \rightarrow \infty}\left[\frac{2}{2-\frac{1}{n^{2}}} \cdot \frac{1}{n} \cos \left(\frac{1+1 / n}{2-1 / n}\right)-\frac{1}{\left(\frac{1}{n}-2\right)} \cdot \frac{1}{\left(1+\frac{1}{n^{2}}\right)} \cdot \frac{1}{n}\right]\)

Given \(f(x)=x^{2}-\pi^{2}\) \(\lim _{x \rightarrow-\pi} \frac{x^{2}-\pi^{2}}{\sin (\sin x)}=\lim _{h \rightarrow 0} \frac{(-\pi+h)^{2}-\pi^{2}}{\sin (\sin (-\pi+h))}\)

\(\lim _{x \rightarrow 1} \frac{x \sin (x-[x])}{x-1}\)

\(\lim _{x \rightarrow-\infty} \frac{x^{2} \tan \frac{1}{x}}{\sqrt{8 x^{2}+7 x+1}}=\lim _{x \rightarrow-\infty} \frac{x^{2} \tan \frac{1}{x}}{-x \sqrt{8+\frac{7}{x}+\frac{1}{x^{2}}}}\)

$$\lim _{x \rightarrow 0} \frac{x^{a} \sin ^{b} x}{\sin x^{c}}=\lim _{x \rightarrow 0} x^{a}\left(\frac{\sin x}{x}\right)^{b}\left(\frac{x^{c}}{\sin x^{c}}\right) x^{b-c}=\lim _{x \rightarrow 0} x^{a+b-c}$$ This limit will have non-zero value if \(a+b=c\).

\(\frac{x^{4}\left(\cot ^{4} x-\cot ^{2} x+1\right)}{\left(\tan ^{4} x-\tan ^{2} x+1\right)}\)

We know that \(\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)=\left\\{\begin{array}{l}2 \tan ^{-1} x, x \geq 0 \\ -2 \tan ^{-1} x, x \leq 0\end{array}\right.\)

\(\lim _{x \rightarrow 1} \frac{1+\sin \pi\left(\frac{3 x}{1+x^{2}}\right)}{1+\cos \pi x}=\lim _{x \rightarrow 1} \frac{1-\cos \left(\frac{3 \pi}{2}-\frac{3 \pi x}{1+x^{2}}\right)}{1-\cos (\pi-\pi x)}\)

\(\lim _{x \rightarrow-1}\left(\frac{x^{4}+x^{2}+x+1}{x^{2}-x+1}\right)^{\lim _{x \rightarrow-1} \frac{1-\cos (x+1)}{(x+1)^{2}}}\)

Given limit is $$ \lim _{x \rightarrow \infty}(x+1)\left[\tan ^{-1}(x+5)-(x+1)\right]+4 \tan ^{-1}(x+5) $$

\(\lim _{x \rightarrow x} \frac{1}{x+1} \tan \left(\frac{\pi x+1}{2 x+2}\right)=\lim _{x \rightarrow \infty} \frac{1}{x+1} \cot \left(\frac{\pi}{2}-\frac{\pi x+1}{2 x+2}\right)\)

\(L=\lim _{x \rightarrow \pi / 2} \frac{\sin (x \cos x)}{\sin \left(\frac{\pi}{2}-x \sin x\right)}\)

\(\lim _{x \rightarrow 0^{+}}\left[\left(1-e^{x}\right) \frac{\sin x}{|x|}\right]=\lim _{x \rightarrow 0^{+}}\left[\left(0^{-}\right) \frac{\sin x}{x}\right]=\left[0^{-}\right]=-1\)

\(\lim _{x \rightarrow 0} \frac{x\left(e^{x}-1\right)}{1-\cos x}=\lim _{x \rightarrow 0} \frac{2 x\left(e^{x}-1\right)}{4 \sin ^{2} \frac{x}{2}}\)

\(\lim _{x \rightarrow 0} \frac{\sin \left(x^{2}\right)}{\ln \left(\cos \left(2 x^{2}-x\right)\right)}\)

\(\lim _{x \rightarrow \infty} \frac{e^{1 / x^{2}}-1}{2 \tan ^{-1}\left(x^{2}\right)-\pi}=\lim _{t \rightarrow 0^{+}} \frac{e^{t^{2}}-1}{2 \cot ^{-1} t^{2}-\pi}\)

\(\lim _{x \rightarrow 0} \frac{\left(2^{m}+x\right)^{1 / m}-\left(2^{n}+x\right)^{1 / n}}{x}\)

\(\lim _{n \rightarrow \infty}\left[\frac{1}{n}+\frac{e^{1 / n}}{n}+\frac{e^{2 / n}}{n}+\cdots+\frac{e^{(n-1) / n}}{n}\right]\)

\(\lim _{x \rightarrow 0} \frac{\log \left(1+x+x^{2}\right)+\log \left(1-x+x^{2}\right)}{\sec x-\cos x}\)

\(\lim _{x \rightarrow \infty}\left(\sqrt[3]{x^{3}+2 x^{2}}-\sqrt{x^{2}+x}\right)\)

\(\lim _{x \rightarrow 0} \frac{1+\sin x-\cos x+\log (1-x)}{x^{3}}\)

\(\cos x=2 \sin \left(\frac{x+\tan x}{2}\right) \sin \left(\frac{x-\tan x}{2}\right)\)

\(\lim _{x \rightarrow 0}\left(\frac{\sin 3 x}{x^{3}}+\frac{a}{x^{2}}+b\right)=\lim _{x \rightarrow 0} \frac{\sin 3 x+a x+b x^{3}}{x^{3}}\)

\(\lim _{x \rightarrow 0} \frac{x^{n} \sin ^{n} x}{x^{n}-\sin ^{n} x}=\lim _{x \rightarrow 0} \frac{x^{n}\left(x-\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}-\cdots\right)^{n}}{x^{n}-\left(x-\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}-\cdots\right)^{n}}\)

\(\lim _{x \rightarrow 0}\left(\frac{1+\tan x}{1+\sin x}\right)^{\operatorname{cosec} x}=\lim _{x \rightarrow 0} \frac{(1+\tan x)^{\frac{1}{\sin x}}}{(1+\sin x)^{\frac{1}{\sin x}}}\)

\(\lim _{m \rightarrow \infty}\left(\cos \frac{x}{m}\right)^{m}=\lim _{m \rightarrow \infty}\left[1-\left(1-\cos \frac{x}{m}\right)\right]^{m}\)

\(\lim _{n \rightarrow \infty}\left(\frac{n^{2}-n+1}{n^{2}-n-1}\right)^{n(n-1)}=\lim _{n \rightarrow \infty}\left(\frac{n(n-1)+1}{n(n-1)-1}\right)^{n(n-1)}\)

\(\lim _{x \rightarrow 0}\left(\frac{1^{x}+2^{x}+\cdots+n^{x}}{n}\right)^{1 / x}=e^{\lim _{x \rightarrow 0}\left(\frac{1^{x}-1}{n}+\frac{2^{x}-1}{n}+\cdots+\frac{n^{x}-1}{n}\right) \frac{1}{x}}\)

\(\lim _{x \rightarrow \infty} \frac{(\log x)^{3}+x \cdot 3(\log x)^{2} \times \frac{1}{x}}{1+2 x}\)

\(L=\lim _{n \rightarrow \infty} \frac{e^{n}}{\left(1+\frac{1}{n}\right)^{n^{2}}}\)