Question

If $f\left(x\right)=\{\begin{array}{cc}m{x}^2+n,& x<0\\ nx+m,& 0\le x\le 1\\ n{x}^3+m,& x>1\end{array}$. For what integers $m$ and $n$ does $\underset{x\to 0}{lim}f\left(x\right)$ and $\underset{x\to 1}{lim}f\left(x\right)$ exist?

Answer 1

The given function is
$f\left(x\right)=\{\begin{array}{cc}m{x}^2+n,& x<0\\ nx+m,& 0\le x\le 1\\ n{x}^3+m,& x>1\end{array}$
$\underset{x\to 0^{-}}{lim}f\left(x\right)$ = $\underset{x\to 0}{lim}(m{x}^2+n)$
$=m{\left(0\right)}^2+n=n$
$\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0}{lim}(nx+m)$
$=n\left(0\right)+m=m$
Thus $\underset{x\to 0}{lim}f\left(x\right)$ exists if $m=n$
$\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1}{lim}(nx+m)$
$=n\left(1\right)+m=m+n$
$\underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{x\to 1}{lim}(n{x}^3+m)$
$=n{\left(1\right)}^3+m$
$=m+n$
$\therefore \underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{x\to 1}{lim}f\left(x\right)$
Thus $\underset{x\to 1}{lim}f\left(x\right)$ exists for any integral value of $m$ and $n$