Differentiation (Part-38) | S N De

 

$\,(xi)\lim_{x  \to a}\,\frac{\sin x-\sin a}{x-a}=\cos a$

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

$\,(xii)\,\lim_{x  \to a}\,\frac{\tan x-\tan a}{x-a}=\sec^2a$

Sol. $\,\,\lim_{x  \to a}\,\frac{\tan x-\tan a}{x-a}\\=\lim_{x  \to a}\frac{\frac{\sin x}{\cos x}-\frac{\sin a}{\cos a}}{x-a}\\=\lim_{x  \to a}\frac{\sin x \cos a-\cos x\sin a}{(x-a) \cos x \cos a}\\=\lim_{x  \to a}\frac{\sin(x-a)}{(x-a)\cos x \cos a}\\=\lim_{x  \to a}\frac{\sin(x-a)}{x-a} \times \lim_{x  \to a}\frac{1}{\cos x \cos a}\\~~[\text{let}\,\,(x-a)=z,\,\,\text{so as}~~ x \to a, z \to 0.]\\=\lim_{z  \to 0}\frac{\sin z}{z} \times \frac{1}{\cos a \cos a}\\=1 \times \frac{1}{\cos^2a}\\=\sec^2a\,\,\text{(proved)}$

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

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

$\,(xiv)\lim_{x  \to 1}\,\frac{\cos(\pi x/2)}{1-x}=\frac{\pi}{2}$

Sol. Let $\,L=\lim_{x  \to 1}\,\frac{\cos(\pi x/2)}{1-x}\rightarrow(1)$

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

So, $\,L=\lim_{z  \to 0}\frac{\cos\left(\frac{\pi}{2}-\frac{\pi z}{2}\right)}{z}\\~~~~= \frac{\pi}{2} \times \lim_{z  \to 0}\frac{\sin(\pi z/2)}{(\pi z /2)}\\=\frac{\pi}{2} \times \lim_{p  \to 0} \frac{\sin p}{p}\\~~~~[\text{let}\,\,\frac{\pi z}{2}=p,\,\,\text{so as}~~ z \to 0, p \to 0.]\\=\frac{\pi}{2} \times 1\\=\frac{\pi}{2}\,\,\text{(proved)}$

$\,(xv)\lim_{\theta \to \frac{\pi}{4}}\,\frac{2-\csc^2\theta}{1-\cot \theta}=2$

Sol. $\,~~~\lim_{\theta \to \frac{\pi}{4}}\frac{2-\csc^2\theta}{1-\cot\theta}\\=\lim_{\theta \to \frac{\pi}{4}}\frac{2-(1+\cot^2\theta)}{1-\cot\theta}\\=\lim_{\theta \to \frac{\pi}{4}}\frac{1-\cot^2\theta}{1-\cot\theta}\\=\lim_{\theta \to \frac{\pi}{4}}\frac{(1+\cot\theta)(1-\cot\theta)}{(1-\cot\theta)}\\=\lim_{\theta \to \frac{\pi}{4}}(1+\cot\theta)\\=1+1\\=2\,\,\text{(proved)}$

$\,(xvi)\lim_{x  \to 0}\,\frac{\tan2x-x}{3x-\sin x}=\frac 12$

Sol. $\,\lim_{x  \to 0}\,\frac{\tan2x-x}{3x-\sin x}\\=\lim_{x  \to 0}\frac{\frac 1x\left(\tan2x-x\right)}{\frac 1x\left(3x-\sin x\right)}\\=\lim_{x  \to 0}\frac{2 \times \frac{\tan 2x}{2x}-1}{3-\frac{\sin x}{x}}\\=\frac{2 \times \lim_{(2x)  \to 0}\frac{\tan 2x}{2x}-1}{3-\lim_{x  \to 0}\frac{\sin x}{x}}\\=\frac{2 \times 1-1}{3-1}~~[\text{Using}\,\,\lim_{z \to 0}\frac{\tan z}{z}=1]\\=\frac 12\,\,\text{(proved)}$

Note : $\,\lim_{z \to 0}\frac{\tan z}{z}\\=\lim_{z \to 0}\frac{\sin z}{z} \times \lim_{z \to 0}\frac{1}{\cos z}\\=1 \times \frac 11\\=1$

$\,12.\,$ Evaluate :

$\,(i)~~\lim_{h  \to 0}\frac{e^{(x+h)^2}-e^{x^2}}{h}\\=\lim_{h  \to 0}\frac{e^{x^2+2xh+h^2}-e^{x^2}}{h}\\=\lim_{h  \to 0}\frac{e^{x^2}[e^{h(2x+h)}-1]}{h(2x+h)} \times \lim_{h  \to 0}(2x+h)\\=e^{x^2} \times \lim_{z  \to 0}\frac{e^z-1}{z} \times (2x)\\~~~[\text{let}\,\,z=h(2x+h),\,\,\text{so that as}~~h \to 0,\,z \to 0.]\\=2xe^{x^2} \times 1\\=2xe^{x^2}\,\,\text{(ans.)}$

$\,(ii)~~\lim_{x  \to 0}\frac{e^{2x}-1}{\sin3x}\\=\lim_{x  \to 0}\frac{\frac{e^{2x}-1}{2x} \times 2}{\frac{\sin3x}{3x} \times 3}\\=\frac{\lim_{(2x)  \to 0}\frac{e^{2x}-1}{2x} \times 2}{\lim_{(3x)  \to 0}\frac{\sin3x}{3x} \times 3}\\=\frac{1 \times 2}{1 \times 3}~~[\text{Using}~~\lim_{z \to 0}\frac{e^z-1}{z}=1]\\=\frac 23\,\,\text{(ans.)}$

$\,(iii)~\lim_{x  \to 0}~\frac{4^x-1}{\sin2x}\\=\lim_{x  \to 0}\frac{\frac{4^x-1}{x}}{\frac{\sin2x}{2x}\times 2}\\=\frac 12 \times \frac{\lim_{x  \to 0}\frac{4^x-1}{x}}{\lim_{(2x)  \to 0}\frac{\sin2x}{2x}}\\=\frac 12 \times \frac{\log_e 4}{1}~~[\because \lim_{x  \to 0}\frac{a^x-1}{x}=\log_e a]\\=\frac 12 \times \log_e (2^2)\\=\frac 12 \times 2\log_e 2\\=\log_e 2\,\,\text{(ans.)}$

$\,(iv)~~\lim_{h  \to 0}\frac{e^{2h}-1}{e^{3h}-1}\\=\lim_{h  \to 0}\frac{\frac{e^{2h}-1}{2h} \times 2}{\frac{e^{3h}-1}{3h} \times 3}\\=\frac 23 \times \frac{\lim_{(2h) \to 0}\frac{e^{2h}-1}{2h}}{\lim_{(3h)  \to 0}\frac{e^{3h}-1}{3h}}\\=\frac 23 \times \frac 11~~[\because~~\frac{e^z-1}{z} \to 1~~\text{as}~~z \to 0]\\=\frac 23\,\,\text{(ans.)}$

$\,(v)\,\lim_{x  \to 0}\frac{a^{\alpha x}-b^{\beta x}}{x}\\=\lim_{x  \to 0}\frac{(a^{\alpha x}-1)-(b^{\beta x}-1)}{x}\\=\lim_{(\alpha x)  \to 0}\left(\frac{a^{\alpha x}-1}{\alpha x}\right) \times \alpha\\~~~-\lim_{(\beta x)  \to 0}\left(\frac{b^{\beta x}-1}{\beta x}\right) \times \beta\\=\alpha \log_e a-\beta \log_e b\,\,[*]\text{(ans.)}$

Note[*]: $\,\lim_{z \to 0}\frac{a^z-1}{z}=\log_e a$

$\,(vi)\lim_{x  \to 0}\,\frac{\log(1+3x)}{x}\\=\lim_{x  \to 0}\frac{\log(1+3x)}{3x} \times 3\\=\lim_{z \to 0}\frac{\log(1+z)}{z} \times 3\\~~~[\text{let}\,\,z=3x,\,\,\text{and so as}~~ x \to 0, z \to 0.]\\=1 \times 3\\=3\,\,\text{(ans.)}$

$\,(vii)\lim_{x  \to 0}\,\frac{2^x-1}{\sqrt{1+x}-1}\\=\lim_{x  \to 0}\frac{(2^x-1)(\sqrt{1+x}+1)}{(\sqrt{1+x}-1)(\sqrt{1+x}+1)}\\=\lim_{x  \to 0}\frac{(2^x-1)((\sqrt{1+x}+1))}{(\sqrt{1+x})^2-1^2}\\=\lim_{x  \to 0}\frac{(2^x-1)(\sqrt{1+x}+1)}{1+x-1}\\=\lim_{x  \to 0}\frac{2^x-1}{x} \times \lim_{x  \to 0}(\sqrt{1+x}+1)\\=\log_e 2 \times (\sqrt{1+0}+1)\\=2\log_e 2\,\,\text{(ans.)}$

$\,(viii)\lim_{x  \to 0}\,\frac{e^x+e^{-x}-2}{x^2}\\=\lim_{x  \to 0}\frac{e^x(e^x+e^{-x}-2)}{x^2e^x}\\=\lim_{x  \to 0}\frac{e^{2x}+1-2e^x}{x^2e^x}\\=\lim_{x  \to 0}\frac{(e^x)^2-2e^x\cdot1+1^2}{x^2e^x}\\=\lim_{x  \to 0}\frac{(e^x-1)^2}{x^2e^x}\\=\lim_{x  \to 0}\left[\left(\frac{e^x-1}{x}\right)^2 \times \frac{1}{e^x}\right]\\=\left(\lim_{x  \to 0}\frac{e^x-1}{x}\right)^2 \times \lim_{x  \to 0} \frac{1}{e^x}\\=1^2 \times \frac{1}{e^0}\\=1 \times \frac 11\\=1\,\,\text{(ans.)}$