Differentiation (Part-38) | S N De

 

$\,15.\,$ Evaluate the following limits :

$\,(i)\,\lim_{x \to 0}\frac{\sin 2x}{2x^2+x}\\=\lim_{x \to 0}\frac{\sin2x}{x(2x+1)}\\=\lim_{2x \to 0}\left(\frac{\sin 2x}{2x}\right)\times \lim_{x \to 0}\frac{2}{2x+1}~~[*]\\=1 \times \frac{2}{2 \times 0+1}\\=2\,\,\text{(ans.)}$

Note[*] : When $\,x \to 0,\, 2x \to 0.$

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

$\,(iii)\,\lim_{x \to 0}\frac{\sin3x}{\sin2x}\\=\lim_{x \to 0}\frac{\frac{\sin3x}{3x} \times 3}{\frac{\sin2x}{2x} \times 2}\\=\frac 32 \times \frac{\lim_{z \to 0}\frac{\sin z}{z}}{\lim_{p \to 0}\frac{\sin p}{p}}\\~~[\text{let}\,z=3x,\,p=2x\,\,\text{so that as}~~x \to 0, \\~~~ \, p\to 0, z \to 0.]\\=\frac 32 \times \frac 11 \\=\frac 32\,\,\text{(ans.)}$

$\,(iv)\,\lim_{x \to \pi}\frac{\sin x}{\pi-x}\\=\lim_{z \to 0}\frac{\sin z}{z}\\~~[\text{let}\,\,z=\pi-x,\,\,\text{so that as}\,\,x \to \pi,z \to 0]\\=\lim_{z \to 0}\frac{\sin z}{z}\\=1\,\,\text{(ans.)}$

$\,(v)\,\lim_{x \to \pi/2}\frac{2x-\pi}{\cos x}\\=\lim_{x \to \pi/2}\frac{-2(\pi/2-x)}{\sin(\pi/2-x)}\\=-2 \times \frac{1}{\lim_{z \to 0}\frac{\sin z}{z}}~~[*]\\= -2 \times \frac 11 \\=-2\,\,\text{(ans.)}$

Note[*] : Let $\,\,z=\pi/2-x,\,\text{so that as}\,\, x \to \pi/2, z \to 0.$

$\,(vi)\,\lim_{x \to 0}\,\frac{\sin x^{\circ}}{x}\\=\lim_{x \to 0}\frac{\sin\frac{\pi x}{180}}{\frac{\pi x}{180}} \times \frac{\pi}{180}\\=\lim_{z \to 0}\frac{\sin z}{z} \times \frac{\pi}{180}~~[*]\\=1 \times \frac{\pi}{180}\\=\frac{\pi}{180}\,\,\text{(ans.)}$

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

$\,16.\,$ Evaluate :

$\,(i)\,\lim_{x \to 0}\,\frac{e^{4x}-1}{x}\\=\lim_{x \to 0}\frac{e^{4x}-1}{4x} \times 4\\=\lim_{z \to 0}\frac{e^z-1}{z} \times 4~~[*]\\=1\times 4\\=4\,\,\text{(ans.)}$

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

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

$\,(iii)\lim_{x \to 0}\,\frac{e^{13x}-e^{7x}}{x}$

Sol. $\,\,~\lim_{x \to 0}~\frac{e^{13x}-e^{7x}}{x}\\=\lim_{x \to 0}\frac{(e^{13x}-1)-(e^{7x}-1)}{x}\\=\lim_{x \to 0}\frac{e^{13x}-1}{x}-\lim_{x \to 0}\frac{e^{7x}-1}{x}\\=\lim_{(13x) \to 0}\frac{e^{13x}-1}{13x}\times 13-\lim_{(7x) \to 0}\frac{e^{7x}-1}{7x} \times 7\\=1 \times 13-1 \times 7\\=13-7\\=6\,\,\text{(ans.)}$

$\,(iv)\,\lim_{x \to 0}\frac{e^{\alpha x}+e^{\beta x}-2}{x}$

Sol. $\,~~~\lim_{x \to 0}~\frac{e^{\alpha x}+e^{\beta x}-2}{x}\\=\lim_{x \to 0}\frac{(e^{\alpha x}-1)+(e^{\beta x}-1)}{x}\\=\lim_{(\alpha x) \to 0}\frac{e^{\alpha x}-1}{\alpha x} \times \alpha+\lim_{(\beta x) \to 0}\frac{e^{\beta x}-1}{\beta x} \times \beta\\=1 \times \alpha +1 \times \beta\\=\alpha+\beta\,\,\text{(ans.)}$

$\,(v)\,\lim_{x \to 0}\frac{e^{\sin x}-1}{\sin x}$

Sol. $\,~~\lim_{x \to 0}\frac{e^{\sin x}-1}{\sin x}\\=\lim_{x \to 0}\frac{e^{\sin x}-1}{\sin x} \times \lim_{x \to 0}\frac{\sin x}{x}\\=\lim_{z \to 0}\frac{e^z-1}{z} \times 1 ~~[*]\\=1 \times 1\\=1\,\,\text{(ans.)}$

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

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

Sol. $\,~~~\lim_{h \to 0}\frac{e^{\tan h}-1}{h}\\=\lim_{h \to 0}\frac{e^{\tan h}-1}{\tan h} \times \lim_{h \to 0}\frac{\tan h}{h}\\=\lim_{z \to 0}\frac{e^z-1}{z} \times \lim_{h \to 0}\frac{\sin h}{h} \times \frac{1}{\lim_{h \to 0}(\cos h)}~~[*]\\=1 \times 1 \times \frac{1}{\cos 0}\\=1 \times 1 \times \frac 11\\=1\,\,\text{(ans.)}$

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