Question

Is the function defined by $f\left(x\right)=x^2-\sin x+5$ continuous at $x=\pi$?

Enter '0' if yes else enter '1.'

Answer 1

The given function is $f\left(x\right)=x^2-\sin x+5$

At $x=\pi ,f\left(x\right)=f\left(\pi \right)={\pi }^2-\sin \pi +5={\pi }^2-0+5={\pi }^2+5$

and $\underset{x\to \pi }{lim}f\left(x\right)=\underset{x\to \pi }{lim}(x^2-sinx+5)={\pi }^2-0+5={\pi }^2+5$

Clearly $\underset{x\to \pi }{lim}=f\left(\pi \right)$

Thus f is continuous at $x=\pi$