Differentiation (Part-38) | S N De

 

$\,13.\,$ Prove :

$\,(i)\,\lim_{x  \to 0}\frac{\log(1+2x)}{\sin3x}=\frac 23$

Sol. $\,\lim_{x  \to 0}\frac{\log(1+2x)}{\sin3x}\\=\lim_{x  \to 0}\frac{\frac{\log(1+2x)}{2x} \times 2}{\frac{\sin3x}{3x}\times 3}\\=\frac{2}{3} \times \frac{\lim_{z  \to 0}\frac{\log(1+z)}{z}}{\lim_{p  \to 0}\frac{\sin p}{p}}\\~~[\text{let}~~z=2x,\,p=3x \\ \,\,\text{so that as}~~x \to 0, z \to 0, p\to 0.]\\=\frac 23 \times \frac 11 \\=\frac 23\,\,\text{(proved)}$

$\,(ii)\,\lim_{x  \to 1}\frac{\log x}{x-1}=1$

Sol. Let $~~L=\lim_{x  \to 1}\frac{\log x}{x-1} \rightarrow(1)$

Let $\,z=x-1\,\,\text{so that as}\,\, x \to 1,\,z \to 0.$

Also, $\,z=x-1 \Rightarrow x=1+z$

So, $\,L=\lim_{z \to 0}\frac{\log(1+z)}{z}~~~~[\text{By (1)}]\\~~~~=1\,\,\text{(proved)}$

$\,(iii)\,\lim_{x  \to e}\frac{\log x-1}{x-e}=\frac 1e$

Sol. $~\lim_{x  \to e}~\frac{\log x-1}{x-e}\\=\lim_{p  \to 0}\frac{\log(p+e)-\log e}{p}\\~~[\text{let}~~x-e=p,\,\,\text{so that as}~~x \to e, p \to 0.]\\=\lim_{p  \to 0}\frac{\log\left(\frac{p+e}{e}\right)}{p}\\=\lim_{p  \to 0}\frac{\log \left(1+\frac pe\right)}{\frac pe \times e}\\=\frac 1e \times \lim_{z  \to 0} \frac{\log(1+z)}{z}\\~~[\text{where}\,\,z=\frac pe,\,\text{and as}\,\, p \to 0, z \to 0.]\\=\frac 1e \times 1\\=\frac 1e\,\,\text{(proved)}$

$\,(iv)\lim_{x \to 0}\,\frac{p^x-q^x}{\tan x}=\log(p/q)$

Sol. $\,~~~\lim_{x \to 0}\frac{p^x-q^x}{\tan x}\\=\lim_{x \to 0}\frac{q^x\left[(p/q)^x-1\right]}{\tan x}\\=\lim_{x \to 0}q^x \times \lim_{x \to 0} \frac{(p/q)^x-1}{x} \times \lim_{x \to 0}\frac{x}{\tan x}\\=q^0 \times \log(p/q) \times \lim_{x \to 0}\frac{x}{\sin x} \times \lim_{x \to 0}\cos x\\=1 \cdot \log(p/q) \times \frac{1}{\lim_{x \to 0}\frac{\sin x}{x}} \times 1\\=\log(p/q) \times \frac 11\\=\log(p/q)\,\,\text{(proved)}$

$\,(v)\,\lim_{x \to 0}\frac{\sin\log(1+x)}{x}=1$

Sol. Let $\,L=\lim_{x \to 0}\frac{\sin\log(1+x)}{x} \rightarrow(1)$

Suppose, $\,z=\log(1+x),\,\,\text{so  as}\,\, x \to 0, z \to \log(1)=0.$

Also, $\,z=\log(1+x)\\ \Rightarrow 1+x=e^z \\\Rightarrow x=e^z-1 \rightarrow(2)$

So, $\,L=\lim_{z  \to 0}\frac{\sin z}{e^z-1}~~[\text{By (1),(2)}]\\~~~=\lim_{z  \to 0}\frac{\sin z}{z} \times \lim_{z  \to 0}\frac{z}{e^z-1}\\~~~=1 \times \frac{1}{\lim_{z  \to 0}\frac{e^z-1}{z}}\\~~~=1 \times \frac 11\\~~~=1\,\,\text{(proved)}$

$\,(vi)\,\lim_{x \to 0}\frac{\log(1+x)+\sin x}{e^x-1}=2$

Sol. $\,\lim_{x \to 0}\frac{\log(1+x)+\sin x}{e^x-1}\\=\frac{\lim_{x \to 0}\frac{\log(1+x)}{x}+\lim_{x \to 0}\left(\frac{\sin x}{x}\right)}{\lim_{x \to 0}\left(\frac{e^x-1}{x}\right)}\\=\frac{1+1}{1}\\=\frac 21\\=2\,\,\text{(proved)}$

$\,(vii)\,\lim_{x \to 0}\frac{\log(\cos x)}{\sin^2x}=-\frac 12.$

Sol. $\,~~~\frac{\log(\cos x)}{\sin^2x}\\=\lim_{x \to 0}\frac{\log[1+(\cos x-1)]}{\sin^2x}\\=\lim_{x \to 0}\frac{\log[1+(\cos x-1)]}{(\cos x-1)} \times \frac{(\cos x-1)}{\sin^2x}\\=\lim_{z \to 0}\frac{\log(1+z)}{z} \times \lim_{x \to 0}\frac{-(1-\cos x)}{(2\sin(x/2)\cos(x/2))^2}~~~[*]\\=1 \times \lim_{x \to 0}\frac{-2\sin^2(x/2)}{4\sin^2(x/2)\cos^2(x/2)}\\=-\frac 12 \times  \lim_{x \to 0}\frac{1}{\cos^2(x/2)}\\=-\frac 12 \times \frac{1}{1^2}\\=-\frac 12\,\,\text{(proved)}$

Note[*] : Let $\,z=\cos x-1,\,\text{so as}\,\, x \to 0,\, z \to 0.$

$\,(viii)~\lim_{h \to  0}~\frac{\log(x+h)-\log x}{h}=\frac 1x$

Sol. $\,\lim_{h \to  0}\frac{\log(x+h)-\log x}{h}\\=\lim_{h \to  0}\frac{\log\left(\frac{x+h}{x}\right)}{h/x} \times \frac 1x\\=\frac 1x \times  \lim_{h \to  0}\frac{\log[1+(h/x)]}{h/x}\\=\frac 1x \times \lim_{z \to  0}\frac{\log(1+z)}{z}~~[*]\\=\frac 1x \times 1 \\=\frac 1x\,\,\text{(proved)}$

Note[*] : Let $\,z=1+(h/x),\,\,\text{so as}\,\, h \to 0, z \to 0.$

$\,(ix)\,\lim_{x \to 0}\frac{\log(1+\sin x)}{x}=1$

Sol. $\,\,~\lim_{x  \to 0}~~\frac{\log(1+\sin x)}{x}\\=\lim_{x \to 0}\frac{\log(1+\sin x)}{\sin x} \times \lim_{x \to 0}\frac{\sin x}{x}\\=\lim_{z  \to 0}\frac{\log(1+z)}{z} \times 1~~[*]\\=1\times 1\\=1\,\,\text{(proved)}$

Note[*] : Let $\,z=\sin x\,\,\text{so that as }\, x \to 0,\,z \to 0.$

$\,(x)\,\lim_{x  \to 0}\,\frac{\sqrt{1+x}-1}{\log(1+x)}=\frac 12$

Sol. $\,\,~~\lim_{x  \to 0}\,\frac{\sqrt{1+x}-1}{\log(1+x)}\\=\lim_{x  \to 0}\frac{(\sqrt{1+x}-1)(\sqrt{1+x}+1)}{(\sqrt{1+x}+1)\log(1+x)}\\=\lim_{x  \to 0}\frac{(\sqrt{1+x})^2-1^2}{\log(1+x)} \times \lim_{x  \to 0}\frac{1}{\sqrt{1+x}+1}\\=\lim_{x  \to 0}\frac{1+x-1}{\log(1+x)} \times  \frac{1}{\sqrt{1+0}+1}\\=\lim_{x  \to 0}\frac{x}{\log(1+x)} \times \frac 12 \\=\frac{1}{\lim_{x  \to 0}\frac{\log(1+x)}{x}} \times\frac 12\\=\frac 11 \times \frac 12\\=\frac 12\,\,\text{(proved)}$