Differentiation (Part-38) | S N De

 

$\,11.\,$ Prove :

$\,(i)\,\, \lim_{x \to 0}\frac{\sin \alpha x^{\circ}}{\sin\beta x^{\circ}}=\frac{\alpha}{\beta}$

Sol. $\,~~~\lim_{x \to 0}\frac{\sin \alpha x^{\circ}}{\sin\beta x^{\circ}}\\=\lim_{x \to 0}\frac{\sin\frac{\pi \alpha x}{180}}{\sin\frac{\pi \beta x}{180}}\\=\lim_{x \to 0} \left(\frac{\sin\frac{\pi \alpha x}{180}}{\frac{\pi \alpha x}{180}} \times \frac{\pi \alpha x}{180} \right)\\~~~\times \frac{1}{\lim_{x \to 0}\left(\frac{\sin\frac{\pi \beta x}{180}}{\frac{\pi \beta x}{180}} \times \frac{\pi \beta x}{180}\right)}\\=\lim_{x \to 0}\frac{\frac{\pi \alpha x}{180}}{\frac{\pi \beta x}{180}} \times \left(\lim_{z \to 0}~~\frac{\sin z}{z}\right)\times \frac{1}{\left(\lim_{p \to 0} \frac{\sin p}{p}\right)}~~[*]\\=\frac{\alpha}{\beta} \times 1 \times \frac 11\\=\frac{\alpha}{\beta}\,\,\text{(proved.)}$

Note[*]: Let $\,z=\frac{\pi \alpha x}{180},\,\text{as}~~ x \to 0,\,z \to 0.$

 $~~~~~~~~~~~~~~\,p=\frac{\pi \beta x}{180},\,\text{as}~~ x \to 0,\,p \to 0.$

$\,(ii)\,\,\lim_{x \to 0}\left(\frac{1}{\sin x}-\frac{1}{\tan x}\right)=0$

Sol. $\,\,\lim_{x \to 0}\left(\frac{1}{\sin x}-\frac{1}{\tan x}\right)\\=\lim_{x \to 0}\left(\frac{1}{\sin x}-\frac{\cos x}{\sin x}\right)\\=\lim_{x \to 0}\left(\frac{1-\cos x}{\sin x}\right)\\=\lim_{x \to 0}\frac{2\sin^2(x/2)}{2\sin(x/2)\cos(x/2)}\\=\lim_{x \to 0}\tan(x/2)\\=0\,\,\text{(proved)}$

$\,11(iii)\,\lim_{x \to 0}\,\frac{\tan x-\sin x}{x^3}=\frac 12.$

Sol. $\,\,\lim_{x \to 0}\,\frac{\tan x-\sin x}{x^3}\\=\lim_{x \to 0}\frac{\frac{\sin x}{\cos x}-\sin x}{x^3}\\=\frac{\sin x-\sin x\cos x}{x^3 \cos x}\\=\lim_{x \to 0}\frac{\sin x(1-\cos x)}{x^3 \cos x}\\=\lim_{x \to 0}\frac{\sin x}{x} \times \lim_{x \to 0}\frac{1}{\cos x} \times \lim_{x \to 0}\frac{1-\cos x}{x^2}\\=1 \times  \frac{1}{\cos 0}\times \lim_{x \to 0}\frac{2\sin^2(x/2)}{\frac{x^2}{4} \times 4}\\=1 \times \frac 11 \times  \frac 24 \times \left(\lim_{(x/2) \to 0}\frac{\sin(x/2)}{x/2}\right)^2\\=\frac 12 \times 1^2\\=\frac 12~~~\text{(proved)}$

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

Sol. Let $\,\tan^{-1}x= \theta \Rightarrow \tan\theta=x.\\ \text{So, as}\,\,x \to 0, \theta \to 0.$ 

Again, let Let $\,\sin^{-1}x= \theta_1 \Rightarrow \sin\theta_1=x.\\ \text{So, as}\,\,x \to 0, \theta_1 \to 0.$ 

Now, $\,~~~\lim_{x \to 0}\,\frac{\tan^{-1}x}{\sin^{-1}x}\\=\lim_{x \to 0}\frac{(\tan^{-1}x)/x}{(\sin^{-1}x)/x}\\=\lim_{x \to 0}\frac{\tan^{-1}x}{x} \times \frac{1}{\lim_{x \to 0} \frac{\sin^{-1}x}{x}}\\=\lim_{\theta \to 0}\frac{\theta}{\tan\theta} \times \frac{1}{\lim_{\theta_1 \to 0}\frac{\theta_1}{\sin\theta_1}}\\=\frac{1}{\lim_{\theta \to 0}\frac{\tan\theta}{\theta}} \times \lim_{\theta_1 \to 0}\frac{\sin\theta_1}{\theta_1}\\=\frac 11 \times 1\\=1\,\,\text{(proved)}$

$\,(v)\,\lim_{\theta \to \frac{\pi}{4}}\frac{1-\tan \theta}{1-\cot\theta}=-1$

Sol. $\,\lim_{\theta \to \frac{\pi}{4}}\frac{1-\tan\theta}{1-\cot\theta}\\=\lim_{\theta \to \frac{\pi}{4}}\frac{\tan\theta(1-\tan\theta)}{\tan\theta-1}~~[\because \cot\theta=\frac{1}{\tan\theta}]\\=-1 \times \lim_{\theta \to \frac{\pi}{4}}\frac{\tan\theta(\tan\theta-1)}{(\tan\theta-1)}\\=-1\times \lim_{\theta \to \frac{\pi}{4}}\tan\theta\\=-1\,\,\text{(proved)}$

$\,(vi)\lim_{h \to 0}\,\frac{\sec(x+h)-\sec x}{h}=\sec x\tan x$

Sol. $\,\lim_{h \to 0}\,\frac{\sec(x+h)-\sec x}{h}\\=\lim_{h \to 0}\frac 1h\left[\frac{1}{\cos(x+h)}-\frac{1}{\cos x}\right]\\=\lim_{h \to 0}\frac{\cos x-\cos(x+h)}{h\cos x\cos(x+h)}\\=\lim_{h \to 0}\frac{2\sin\frac{2x+h}{2}\sin\frac h2}{h\cos x\cos(x+h)}\\=\lim_{(h/2) \to 0}\frac{\sin(h/2)}{h/2} \times \lim_{h \to 0}\frac{\sin\frac{2x+h}{2}}{\cos x\cos(x+h)}\\=1 \times \frac{\sin x}{\cos x \cos x}\\=\frac{\sin x}{\cos x} \times \frac{1}{\cos x}\\=\tan x\sec x\,\,\text{(proved)}$

$\,(vii)\,\lim_{x \to 0}\frac{\cos9x-\cos7x}{\cos7x-\cos5x}$

Sol. We know, $\,\cos A-\cos B\\=-2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)\rightarrow(1)$

$\,\lim_{x \to 0}\,\frac{\cos9x-\cos7x}{\cos7x-\cos5x}\\=\lim_{x \to 0}\frac{-2\sin8x\sin x}{-2\sin6x\sin x}\,\,[\text{By (1)}]\\=\lim_{x \to 0}\frac{\sin 8x}{\sin6x}\\=\lim_{8x \to 0}\frac{\sin8x}{8x} \times\lim_{x \to 0}\frac{8x}{\frac{\sin 6x}{6x}\times 6x}\\=\frac 86 \times 1\times \frac{1}{\lim_{6x \to 0}\frac{\sin6x}{6x}}\\=\frac 43 \times \frac 11\\=\frac 43\,\,\text{(proved)}$

$\,(viii)\,\lim_{x \to 0}\frac{\sin6x+\sin8x}{\sin4x+\sin6x}$

Sol. We know, $\,\sin A+\sin B\\=2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)\rightarrow(1)$

$\,\,~\lim_{x \to 0}~\frac{\sin6x+\sin8x}{\sin4x+\sin6x}\\=\lim_{x \to 0}\frac{2\sin7x\cos x}{2\sin5x\cos x}\,\,[\text{By (1)}]\\=\lim_{x \to 0}\frac{\sin7x}{\sin5x}\\=\lim_{x \to 0}\frac{\frac{\sin7x}{7x}\times 7}{\frac{\sin5x}{5x}\times 5}\\=\frac 75 \times  \frac{\lim_{7x \to 0}\frac{\sin7x}{7x}}{\lim_{5x \to 0}\frac{\sin 5x}{5x}}\\=\frac 75 \times \frac 11 \\=\frac 75\,\,\text{(proved)}$

$\,(ix)\,\lim_{\theta \to\frac{\pi}{2}}\,(\sec\theta -\tan\theta )$

Sol. $\,\,\lim_{\theta \to\frac{\pi}{2}}\,(\sec\theta -\tan\theta )\\=\lim_{\theta \to\frac{\pi}{2}}\left(\frac{1}{\cos\theta}-\frac{\sin\theta}{\cos\theta}\right)\\=\lim_{\theta \to\frac{\pi}{2}}\frac{1-\sin\theta}{\cos\theta}\\=\lim_{\theta \to\frac{\pi}{2}}\frac{(1-\sin\theta)(1+\sin\theta)}{\cos\theta(1+\sin\theta)}\\=\lim_{\theta \to\frac{\pi}{2}}\frac{1-\sin^2\theta}{\cos\theta(1+\sin\theta)}\\=\lim_{\theta \to\frac{\pi}{2}}\frac{\cos^2\theta}{\cos\theta(1+\sin\theta)}\\=\lim_{\theta \to\frac{\pi}{2}}\frac{\cos\theta}{1+\sin\theta}\\=\frac 02~~[\because \sin (\frac{\pi}{2})=1,\,\cos(\frac{\pi}{2})=0]\\=0\,\,\text{(proved)}$

$\,(x)\,\lim_{x \to y} \frac{\cos^2x-\cos^2y}{x^2-y^2}=-\frac{\sin2y}{2y}$

Sol. $\,\, \lim_{x \to y}\frac{\cos^2x-\cos^2y}{x^2-y^2}\\=\lim_{x \to y}\frac{(1-\sin^2x)-(1-\sin^2y)}{x^2-y^2}\\=\lim_{x \to y}\frac{-(\sin^2x-\sin^2y)}{x^2-y^2}\\=-1 \times \lim_{x \to y}\frac{\sin(x+y)\sin(x-y)}{(x+y)(x-y)}\\=-1 \times \lim_{x \to y}\frac{\sin(x+y)}{(x+y)} \times \lim_{x \to y}\frac{\sin(x-y)}{x-y}\\=-1\times \frac{\sin 2y}{2y} \times \lim_{z \to 0}\frac{\sin z}{z}~~~[*]\\=-\frac{\sin2y}{2y} \times 1\\=-\frac{\sin2y}{2y}\,\,\text{(proved)}$

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