Question

If $f\left(x\right)=\left\{\begin{array}{ll}\frac{x^2-1}{x-1};\mathrm{for}& x\ne 1\\ 2;\mathrm{for}& x=1\end{array}\right.$
Find whether f(x) is continuous at x = 1.

Answer 1

Given:
$f\left(x\right)=\left\{\begin{array}{l}\frac{x^2-1}{x-1},\mathrm{if}x\ne 1\\ 2,\mathrm{if}x=1\end{array}\right.$

We observe
(LHL at x = 1) = $\underset{\mathrm{x}\to 1^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(1-h\right)$
= ​$\underset{h\to 0}{\lim }\frac{{\left(1-h\right)}^2-1}{\left(1-h\right)-1}=\underset{h\to 0}{\lim }\frac{1+h^2-2h-1}{1-h-1}=\underset{h\to 0}{\lim }\frac{h^2-2h}{-h}=\underset{h\to 0}{\lim }\frac{h\left(h-2\right)}{-h}=\underset{h\to 0}{\lim }\left(2-h\right)=2$

(RHL at x = 1) = $\underset{\mathrm{x}\to 1^{+}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(1+h\right)$
= ​$\underset{h\to 0}{\lim }\frac{{\left(1+h\right)}^2-1}{\left(1+h\right)-1}=\underset{h\to 0}{\lim }\frac{1+h^2+2h-1}{1+h-1}=\underset{h\to 0}{\lim }\frac{h^2+2h}{h}=\underset{h\to 0}{\lim }\frac{h\left(h+2\right)}{h}=\underset{h\to 0}{\lim }\left(2+h\right)=2$

Given:
$f\left(1\right)=2$


$\therefore \underset{\mathrm{x}\to 1^{-}}{\lim }f\left(x\right)=\underset{\mathrm{x}\to 1^{+}}{\lim }f\left(x\right)=f\left(1\right)$

Hence, $f\left(x\right)$ is continuous at $x=1$.