Differentiation (Part-38) | S N De

 $\,10(ix)\,\lim_{h  \to 0}\frac{\sin(x+h)-\sin x}{h}$

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

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

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

$\,(xi)\,\lim_{x  \to 0}\frac{\sin(x^2+4x)}{x^3-5x^2+2x}$

Sol. $\,\,\lim_{x  \to 0}\frac{\sin(x^2+4x)}{x^3-5x^2+2x}\\=\lim_{x  \to 0}\frac{\sin(x^2+4x)}{x^2+4x} \times \lim_{x  \to 0}\frac{x^2+4x}{x^3-5x^2+2x}\\=\lim_{z \to 0}\frac{\sin z}{z} \times \lim_{x  \to 0} \left(\frac{x(x+4)}{x(x^2-5x+2)}\right)~~[*]\\=1 \times \lim_{x  \to 0}\frac{x+4}{x^2-5x+2}\\=\frac{0+4}{0^2-5 \times 0+2}\\=\frac 42\\=2$

Note[*] : Let $\,z=x^2+4x,\,\,\text{and so as}\,\, x \to 0,~~ z \to 0.$

$\,(xii)\,\lim_{x  \to 0}\frac{\sin x(1- \cos x)}{x^3}$

Sol. $\,\,\lim_{x  \to 0}\frac{\sin x(1- \cos x)}{x^3}\\=\lim_{x  \to 0}\frac{\sin x}{x} \times \lim_{x  \to 0}\left(\frac{1-\cos x}{x^2}\right)\\=1 \times \lim_{x  \to 0} \frac{2 \sin^2\frac x2}{x^2}\\=\lim_{x  \to 0}\frac{2\sin^2 (x/2)}{\frac{x^2}{4} \times 4}\\=\frac 24 \times  \left(\lim_{x/2  \to 0}\frac{\sin(x/2)}{x/2}\right)^2\\=\frac 12 \times 1^2\\=\frac 12 \,\,\text{(ans.)}$

$\,(xiii)\, \lim_{x  \to 0}\tan5x\csc4x$

Sol. $\,\, \lim_{x  \to 0}\tan5x\csc4x\\=\lim_{x  \to 0}\left(\frac{\sin5x}{\cos 5x}\times \frac{1}{\sin4x}\right)\\=\lim_{x  \to 0}\left(\frac{\sin5x}{5x} \times 5x \right)\times \left(\lim_{x  \to 0}\frac{1}{\cos 5x}\right)\\~~~~~~\times \frac{1}{\lim_{x  \to 0}\left(\frac{\sin 4x}{4x} \times 4x\right)}\\=\lim_{x  \to 0}\left(\frac{5x}{4x}\right) \times \left(\lim_{5x  \to 0} \frac{\sin5x}{5x}\right) \\~~~~\times \left(\frac{1}{\lim_{5x  \to 0} \cos 5x}\right) \times \frac{1}{\lim_{4x  \to 0}\left(\frac{\sin 4x}{4x} \right)}\\=\frac 54 \times 1 \times  \frac{1}{\cos 0} \times \frac{1}{1}\\=\frac 54\,\,[\because \cos 0=1]\,\,\text{(ans.)}$

$\,(xiv)\,\lim_{x  \to 0}\frac{2\sin x-\sin2x}{x^3}$

Sol. $\,\,\lim_{x  \to 0}\frac{2\sin x-\sin2x}{x^3}\\=\lim_{x  \to 0}\frac{2\sin x-2\sin x\cos x}{x^3}\\=\lim_{x  \to 0}\frac{2\sin x(1-\cos x)}{x^3}\\=2 \times \lim_{x  \to 0}\frac{\sin x}{x} \times \lim_{x  \to 0}\frac{1-\cos x}{x^2}\\=2 \times 1 \times \lim_{x  \to 0}\frac{2\sin^2(x/2)}{\frac{x^2}{4} \times 4}\\=2 \times \frac 24 \times \left( \lim_{x/2  \to 0}\frac{\sin(x/2)}{x/2}\right)^2\\=1 \times 1^2\\=1\,\,\text{(ans.)}$

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

Sol. $\,\,\lim_{x  \to a}\frac{1-\cos(x-a)}{(x-a)^2}\\=\lim_{x  \to a}\frac{2\sin^2\frac{x-a}{2}}{(x-a)^2} \\~~~~~~~~~~~~~[\text{let}\,\,z=\frac{x-a}{2}\,\,\text{so that as}\,\,x \to a, z \to 0]\\=\lim_{z \to 0} \frac{2\sin^2z}{4z^2}\\=\frac 24 \times \left(\lim_{z  \to 0}\frac{\sin z}{z}\right)^2\\~~~~~~~~~~=\frac 12 \times 1^2\\=\frac 12\,\,\text{(ans.)}$

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

Sol. $\,\,\lim_{x  \to 0}\frac{\cos x-\sec x}{x^2}\\=\lim_{x  \to 0}\frac{\cos^2x-1}{x^2 \cos x}\\=-1 \times \lim_{x  \to 0}\frac{1-\cos^2x}{x^2} \times \frac{1}{\lim_{x  \to 0}(\cos x)}\\=-1 \times \left(\lim_{x  \to 0}\frac{\sin x}{x}\right)^2 \times \frac{1}{\cos 0}\\=-1 \times 1^2 \times \frac 11~~[\because \cos 0=1]\\=-1\,\,\text{(ans.)}$

$\,(xvii)\,\lim_{x  \to 0}\frac{1-\cos4x}{1-\cos5x}$

Sol. $\,\,\lim_{x  \to 0}\frac{1-\cos4x}{1-\cos5x}\\=\lim_{x  \to 0}\frac{2\sin^22x}{2\sin^2(5x/2)}\\=\lim_{x  \to 0}\frac{\sin^22x}{\sin^2(5x/2)}\\=\lim_{x  \to 0}\left(\frac{\sin 2x}{2x} \times 2x \right)^2 \times  \frac{1}{\left[\lim_{x  \to 0}\frac{\sin(5x/2)}{5x/2} \times \frac{5x}{2}\right]^2}\\= \lim_{x  \to 0}\frac{4x^2}{(5x/2)^2} \times \left(\lim_{2x  \to 0}\frac{\sin2x}{2x}\right)^2 \\~~~~~~~\times  \frac{1}{\left(\lim_{5x/2  \to 0}\frac{\sin(5x/2)}{(5x/2)}\right)^2}\\=\frac{4}{25/4} \times 1^2 \times  \frac{1}{1^2}\\=\frac{16}{25}\,\,\text{(ans.)}$

$\,(xviii)\,\lim_{x  \to \pi} \frac{1+\cos x}{\pi-x}$

Sol. $\, \lim_{x  \to \pi}\frac{1+\cos x}{\pi-x}\\=\lim_{x  \to \pi}\frac{2\cos^2(x/2)}{\pi-x}\\=\lim_{z \to 0}\frac{2\cos^2\left(\frac{\pi}{2}-\frac z2\right)}{z}\\~~~~~~~~~~~~[\text{let}\,\, z=\pi-x,\,\text{as}\,~~ x \to\pi, \,z \to 0]\\=\lim_{z \to 0}\frac{2\sin^2\frac z2}{z}\\=\left(\lim_{(z/2) \to 0}\frac{\sin(z/2)}{z/2}\right)^2 \times \lim_{z \to 0}\frac z2\\=1^2 \times 0\\=0\,\,\text{(ans.)}$