Question

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

Answer 1

Given,

$f\left(x\right)=\{\begin{array}{cc}\frac{x^2-1}{x-1},& \text{if}x\ne 1\\ 2,& \text{if}x=1\end{array}$

LHL at $x=1$

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

RHL at $x=1$

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

Also,

$f\left(1\right)=2\left(\text{Given}\right)$

Now, since

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

Hence $f\left(x\right)$ is continuous at $x=3$.