Question

Let a function $f\left(x\right)$ be defined by

$f\left(x\right)=\{\begin{array}{ccc}\frac{x^2-4}{x-2}& :& x\ne 2\\ 4& :& x=2\end{array}$.

Then check the continuity of the function $f\left(x\right)$ at $x=2$.

Answer 1

Now we've,

$\underset{x\to 2}{lim}\frac{x^2-4}{x-2}$

$=\underset{x\to 2}{lim}\frac{x^2-2^2}{x-2}$

$=2\times 2$

$=4$.

So, we've $\underset{x\to 2}{lim}f\left(x\right)=f\left(2\right)$. Hence the function $f\left(x\right)$ is continuours at $x=2$.