Question

Evaluate the following one sided limits:

(i)
$\underset{x\to 2^{+}}{\lim }\frac{x-3}{x^2-4}$

(ii)
$\underset{x\to 2^{-}}{\lim }\frac{x-3}{x^2-4}$

(iii)
$\underset{x\to 0^{+}}{\lim }\frac{1}{3x}$

(iv)
$\underset{x\to -8^{+}}{\lim }\frac{2x}{x+8}$

(v)
$\underset{x\to 0^{+}}{\lim }\frac{2}{x^{1/5}}$

(vi)
$\underset{x\to \frac{\pi }{2}}{\lim }\tan x$

(vii)
$\underset{x\to -\pi /2^{+}}{\lim }\sec x$

(viii)
$\underset{x\to 0^{-}}{\lim }\frac{x^2-3x+2}{x^3-2x^2}$

(ix)
$\underset{x\to -2^{+}}{\lim }\frac{x^2-1}{2x+4}$

(x)
$\underset{x\to 0^{-}}{\lim }2-\cot x$

(xi)
$\underset{x\to 0^{-}}{\lim }1+\mathrm{cosec}x$

Answer 1

(i)
$$\begin{gathered} \underset{x\to 2^{+}}{\lim }\left(\frac{x-3}{x^2-4}\right) \\ \mathrm{Let}x=2+h,\mathrm{where}h\to 0. \\ \Rightarrow \underset{h\to 0}{\lim }\left[\frac{\left(2+h\right)-3}{{\left(2+h\right)}^2-2^2}\right] \\ =\underset{h\to 0}{\lim }\left[\frac{-1+h}{\left(2+h-2\right)\left(2+h+2\right)}\right] \\ =-\infty \end{gathered}$$

(ii)
$$\begin{gathered} \underset{x\to 2^{-}}{\lim }\left(\frac{x-3}{x^2-4}\right) \\ \mathrm{Let}x=2-h,\mathrm{where}h\to 0. \\ \Rightarrow \underset{h\to 0}{\lim }\left[\frac{2-h-3}{{\left(2-h\right)}^2-2^2}\right] \\ =\underset{h\to 0}{\lim }\left[\frac{-h-1}{\left(2-h-2\right)\left(2-h+2\right)}\right] \\ =\underset{h\to 0}{\lim }\left[\frac{-h-1}{\left(-h\right)\left(4-h\right)}\right] \\ =\underset{h\to 0}{\lim }\left[\frac{1+h}{h\left(4-h\right)}\right] \\ =\infty \end{gathered}$$

(iii)
$$\begin{gathered} \underset{x\to 0^{+}}{\lim }\left(\frac{1}{3x}\right) \\ \mathrm{Let}x=0+h,\mathrm{where}h\to 0. \\ \Rightarrow \underset{h\to 0}{\lim }\left(\frac{1}{3h}\right)=\infty \end{gathered}$$

(iv)
$$\begin{gathered} \underset{x\to -8^{+}}{\lim }\left(\frac{2x}{x+8}\right) \\ \mathrm{Let}x=-8+h,\mathrm{where}h\to 0. \\ \Rightarrow \underset{h\to 0}{\lim }\left(\frac{2\left(-8+h\right)}{-8+h+8}\right) \\ =\underset{h\to 0}{\lim }\left(\frac{-16+2h}{h}\right) \\ =-\infty \end{gathered}$$

(v)
$$\begin{gathered} \underset{x\to 0^{+}}{\lim }\frac{2}{x^{{\displaystyle \frac{1}{5}}}} \\ \mathrm{Let}x=0+h,\mathrm{where}h\to 0. \\ \Rightarrow \underset{h\to 0}{\lim }\left[\frac{2}{h^{{\displaystyle \frac{1}{5}}}}\right] \\ =\infty \end{gathered}$$

(vi)
$$\begin{gathered} \underset{x\to {\frac{\mathrm{\pi }}{2}}^{-}}{\lim }\left(\tan x\right) \\ \mathrm{Let}x=\frac{\mathrm{\pi }}{2}-h,\mathrm{where}h\to 0. \\ \Rightarrow \underset{h\to 0}{\lim }\left[\tan \left(\frac{\mathrm{\pi }}{2}-h\right)\right] \\ =\underset{h\to 0}{\lim }\left(\cot h\right) \\ =\infty \end{gathered}$$

(vii)
$$\begin{gathered} \underset{x\to -{\frac{\mathrm{\pi }}{2}}^{+}}{\lim }\sec x \\ \mathrm{Let}x=-\frac{\mathrm{\pi }}{2}+h,\mathrm{where}h\to 0. \\ \underset{h\to 0}{\lim }\sec \left(-\frac{\mathrm{\pi }}{2}+h\right) \\ =\underset{h\to 0}{\lim }\sec \left(\frac{\mathrm{\pi }}{2}-h\right)\left[\sec \left(-\mathrm{\theta }\right)=\sec \mathrm{\theta }\right] \\ =\underset{h\to 0}{\lim }\mathrm{cosec}h \\ =\infty \end{gathered}$$

(viii)
$$\begin{gathered} \underset{x\to 0^{-}}{\lim }\left[\frac{x^2-3x+2}{x^3-2x^2}\right] \\ =\underset{x\to 0^{-}}{\lim }\left[\frac{x^2-2x-x+2}{x^2\left(x-1\right)}\right] \\ =\underset{x\to 0^{-}}{\lim }\left[\frac{x\left(x-2\right)-1\left(x-2\right)}{x^2\left(x-2\right)}\right] \\ =\underset{x\to 0^{-}}{\lim }\left[\frac{\left(x-1\right)\left(x-2\right)}{x^2\left(x-2\right)}\right] \\ \mathrm{Let}x=0-h,\mathrm{where}h\to 0. \\ \Rightarrow \underset{h\to 0}{\lim }\left[\frac{\left(-h-1\right)}{{\left(-h\right)}^2}\right] \\ =-\infty \end{gathered}$$

(ix)
$$\begin{gathered} \underset{x\to -2^{+}}{\lim }\left(\frac{x^2-1}{2x+4}\right) \\ \mathrm{Let}x=-2+h,\mathrm{where}h\to 0. \\ \underset{h\to 0}{\lim }\left[\frac{{\left(-2+h\right)}^2-1}{2\left(-2+h\right)+4}\right] \\ =\frac{4-1}{-4+4} \\ =\frac{3}{0} \\ =\infty \end{gathered}$$

(x)
$$\begin{gathered} \underset{x\to 0^{-}}{\lim }\left(2-\cot x\right) \\ \mathrm{Let}x=0-h,\mathrm{where}h\to 0. \\ \underset{h\to 0}{\lim }\left[2-\cot \left(-h\right)\right] \\ =\underset{h\to 0}{\lim }\left(2+\cot h\right) \\ =2+\infty \\ =\infty \end{gathered}$$

(xi)
$$\begin{gathered} \underset{x\to 0^{-}}{\lim }\left[1+\cos ecx\right] \\ \mathrm{Let}x=0-h,\mathrm{where}h\to 0. \\ \underset{h\to 0}{\lim }\left[1+\mathrm{cosec}\left(-h\right)\right] \\ =\underset{h\to 0}{\lim }\left[1-\mathrm{cosec}h\right]\left[\mathrm{cosec}\left(-\theta \right)=-\mathrm{cosec}\theta \right] \\ =1-\infty \\ =-\infty \end{gathered}$$