Differentiation (Part-38) | S N De

 $\,10.\,$ Evaluate the following limits :

$\,(i)\,\lim_{x  \to 0}\frac{\sqrt{\cos x}-\sqrt[3]{\cos x}}{\sin^2x}$

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

Now, $\,\lim_{x  \to 0}\frac{\sqrt{\cos x}-\sqrt[3]{\cos x}}{\sin^2x}\\=\lim_{z  \to 1}\frac{z^{\frac 12}-z^{\frac 13}}{1-z^2}\, \\~~~~[\because \sin^2x=1-\cos^2x=1-z^2]\\=\lim_{z  \to 1}\left[-\frac{(z^{\frac 12}-z^{\frac 13})(z+z^{\frac 12}.z^{\frac 13}+z^{\frac 23})}{(z+1)(z-1)(z+z^{\frac 12}.z^{\frac 13}+z^{\frac 23})}\right]\\=-1 \times \lim_{z  \to 1} \frac{(z^{1/2})^3-(z^{1/3})^3}{(z+1)(\sqrt z+1)(\sqrt z-1)(z+z^{\frac 12}.z^{\frac 13}+z^{\frac 23})}\\=-1 \times \lim_{z  \to 1}\frac{z^{3/2}-z}{(z+1)(\sqrt z+1)(\sqrt z-1)(z+z^{\frac 12}.z^{\frac 13}+z^{\frac 23})}\\=-1\times \lim_{z  \to 1} \frac{z(\sqrt z-1)}{(z+1)(\sqrt z+1)(\sqrt z-1)(z+z^{\frac 12}.z^{\frac 13}+z^{\frac 23})}\\=-1 \times \lim_{z  \to 1}\frac{z}{(z+1)(\sqrt z+1)(z+z^{\frac 56}+z^{\frac 23})}\\=\frac{-1 \times 1}{(1+1)(\sqrt 1+1)(1+1^{5/6}+1^{2/3})}\\=\frac{-1}{2 \times 2 \times 3}\\=-\frac{1}{12}\,\,\text{(ans.)}$

$\,(ii)\,\lim_{x \to 0}\frac{1-\cos x}{x^2}$

Sol. $\,\lim_{x \to 0}\,\frac{1-\cos x}{x^2}\\=\lim_{x \to 0} \frac{2\sin^2{(x/2)}}{x^2}\\=\frac 12 \times \lim_{x \to 0} \frac{4 \sin^2(x/2)}{x^2}\\=\frac 12 \times \lim_{x \to 0} \frac{\sin^2(x/2)}{(x/2)^2}\\=\frac 12 \times \lim_{z \to 0} \frac{\sin^2z}{z^2}\\~~~[\text{let}~~ z=x/2,\,\text{so that as}~~x \to 0, z \to 0]\\=\frac 12 \times \left(\lim_{z \to 0} \frac{\sin z}{z}\right)^2\\=\frac 12 \times 1^2\\=\frac 12\,\,\,\text{(ans.)}$

$\,(iii)\,\,\lim_{h  \to 0}\frac{1-\cos h}{h \sin h}$

Sol. $\,\lim_{h  \to 0}\,\,\frac{1-\cos h}{h \sin h}\\=\lim_{h  \to 0}\frac{(1-\cos h)/h^2}{\frac{h\sin h}{h^2}}\\=\lim_{h  \to 0}\frac{2\sin^2(h/2)}{h^2} \times  \frac{1}{\lim_{h  \to 0}\frac{\sin h}{h}}\\=\frac 12 \times \left(\lim_{(h/2)  \to 0}\frac{\sin (h/2)}{h/2}\right)^2 \times \frac 11\\=\frac 12 \times  1^2\\=\frac 12\,\,\text{(ans.)}$

$\,(iv)\,\frac{1-\cos 4x}{x^2}$

Sol. $\,~~~\lim_{x  \to 0}\,\frac{1-\cos 4x}{x^2}\\=\lim_{x  \to 0}\frac{2\sin^22x}{x^2}\\=8 \times \lim_{x  \to 0}\frac{\sin^22x}{(2x)^2}\\=8 \times \left(\lim_{z  \to 0}\frac{\sin z}{z}\right)^2 \\~~~[\text{let}\,\,z=2x,\,\,\text{so as}\,\, x \to 0, z \to 0]\\=8 \times 1^2\\=8\,\,\text{(ans.)}$

$\,(v)\, \lim_{h  \to 0}\frac{\cos ah-\cos bh}{h^2}$

Sol. $\,\lim_{h  \to 0}\, \frac{\cos ah-\cos bh}{h^2}\\=\lim_{h  \to 0}\frac{2\sin \frac{(a+b)h}{2} \sin\frac{(b-a)h}{2}}{h^2}\\= 2 \times \lim_{h  \to 0}\frac{\sin \frac{(a+b)h}{2}}{(a+b)h/2}\\~~~~~~ \times \lim_{h  \to 0}\frac{\sin \frac{(b-a)h}{2}}{(b-a)h/2} \times \frac{(a+b)(b-a)}{4}\\=2 \times  \left(\lim_{y  \to 0} \frac{\sin y}{y}\right) \times  \left(\lim_{z  \to 0} \frac{\sin z}{z}\right)  \\~~~~~~~~~~~\times \left( \frac{(a+b)(b-a)}{4}\right)~~[*]\\=2 \times 1 \times 1 \times  \frac{b^2-a^2}{4}\\=\frac{b^2-a^2}{2}\,\,\text{(ans.)}$

Note[*]: Let $\,y= \frac{(a+b)h}{2},\,\text{and so as}\,\, h \to 0 , y \to 0.$

and also let $\,z= \frac{(b-a)h}{2},\,\text{and so as}\,\, h \to 0 , z \to 0.$

$\,(vi)\,\lim_{x \to 0}\frac{\tan^{-1}x}{x}$

Sol.  Let $\,L=\,\lim_{x \to 0}\frac{\tan^{-1}x}{x}$

Suppose that $\, \tan^{-1}x=\theta \Rightarrow \tan \theta=x.\\ \text{So as}~~ x \to 0, ~~\theta \to 0.$

Hence, $\,L=\lim_{\theta \to 0} \left(\frac{\theta}{ \tan \theta}\right)\\=\lim_{\theta \to 0}\left(\frac{\theta}{ \sin \theta} \times  \cos \theta \right)\\=\frac{1}{ \lim_{\theta \to 0} \left(\frac{\sin \theta}{\theta}\right)} \times \lim_{\theta \to 0} \cos \theta\\=\frac 11 \times \cos 0\\=1 \times 1\\=1\,\,\text{(ans.)}$

$\,(vii)\,\lim_{x \to 0}\frac{x\tan4x}{1-\cos 4x}$

Sol. $\,~~~\lim_{x \to 0}\,\frac{x\tan4x}{1-\cos 4x}\\=\lim_{x \to 0}\frac{x \sin4x}{1-\cos 4x} \times \lim_{x \to 0}\frac{1}{\cos 4x}\\=\lim_{x \to 0}\frac{2x\sin2x\cos 2x}{2\sin^22x} \times \frac{1}{\cos 0}\\=\lim_{x \to 0}\frac{x \sin2x}{\left(\frac{\sin2x}{2x}\right)^2 \times 4x^2} \times \lim_{x \to 0}~\cos 2x \times \frac 11\\=\frac{1}{\left(\lim_{2x \to 0}\frac{\sin2x}{2x}\right)^2} \times \lim_{2x \to 0}\frac{\sin2x}{2x} \\~~~~~~ \times \frac 12 \times \cos 0 \\=\frac 11 \times 1 \times \frac 12 \times 1\\=\frac 12\,\,\text{(ans.)}$

$\,(viii)\,\lim_{x \to 0}\frac{x\tan x}{1-\cos x}$

Sol. $\,~~~\lim_{x \to 0}\,\frac{x\tan x}{1-\cos x}\\=\lim_{x \to 0}\frac{x \sin x}{1-\cos x} \times \lim_{x \to 0}\frac{1}{\cos x}\\=\lim_{x \to 0}\frac{2x\sin x/2\cos x/2}{2\sin^2\frac x2} \times \frac{1}{\cos 0}\\=\lim_{x \to 0}\frac{x}{\sin\frac x2} \times \frac 11\\=\frac{1}{\lim_{x/2 \to 0}\frac{\sin(x/2)}{x/2} \times \frac 12}\\=\frac{1}{1 \times \frac 12}\\=2\,\,\text{(ans.)}$

Alternative Process :

$\,\lim_{x  \to 0}\frac{x\tan x}{1-\cos x}\\=\lim_{x  \to 0}\frac{x \tan x(1+\cos x)}{(1-\cos x)(1+\cos x)}\\=\lim_{x  \to 0}\frac{x\sin x(1+\cos x)}{\cos x(1-\cos^2x)}\\=\lim_{x  \to 0}\frac{x\sin x(1+\cos x)}{\cos x \times \sin^2x}\\=\lim_{x  \to 0}\frac{x}{\sin x} \times  \lim_{x  \to 0}\frac{1+\cos x}{\cos x}\\=1 \times \frac{1 +\cos 0}{\cos 0}\\=1 \times \frac{1+1}{1}\\=2\,\,\text{(ans.)}$