Question

f(x) = $\{\begin{array}{c}{x}^{10}-1,ifx\le 1\\ x^2,ifx>1\end{array}$

Answer 1

Here, f(x) = $\{\begin{array}{c}{x}^{10}-1,ifx\le 1\\ x^2,ifx>1\end{array}$

for x > 1, f(x) = $x^{10}$ - 1 and x > 1, f(x) = $x^2$ is a polynomial funtion, so f is a continuous in the above interval. Therefore, we have to check the continuity at x = 1.

LHL = $\underset{x\to 1^{-}}{lim}$ f(x) = $\underset{x\to 1^{-}}{lim}$ $x^{10}-3$

Putting x=1-h as $x\to 1^{-}$ when $h\to 0$

$\therefore$ $\underset{h\to 0}{lim}$ $\left[\right(1-h{)}^{10}-1]$ = (1-0)^{10} -1=1-1=0

RHL = $\underset{x\to 1^{+}}{lim}$ f(x) = $\underset{x\to 1^{+}}{lim}$ $\left(x^2\right)$

Putting x=1+h as $x\to 1^{+}$ when $x\to 0$

$\underset{h\to 0}{lim}$ $(2+h{)}^2$ = $\underset{h\to 0}{lim}$ $(1+h^2+2h)$ =1

$\therefore$ LHL = RHL

Thus, f(x) is continuous at x=1.