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 integeres $m$ and $n$ does both $\underset{x\to 0}{lim}f\left(x\right)$ and $\underset{x\to 1}{lim}f\left(x\right)$ exist ?

Answer 1

Step $1:$ Solve $\underset{x\to 0}{lim}f\left(x\right)$
Given $f\left(x\right)\{\begin{array}{cc}m{x}^2+n,& x<0\\ nx+m& 0\le x\le 1\\ n{x}^3+m,& x>1\end{array}$
Limit exists at $x=0$ if
$L.H.L.=R.H.L.$

$L.H.L.=\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(0-h)$
$\Rightarrow L.H.L.=\underset{h\to 0}{lim}f(-h)$
$\Rightarrow L.H.L.=\underset{h\to 0}{lim}\left(m\right(-h^2)+n)$
$\Rightarrow L.H.L.=m(0{)}^2+n=n$

$R.H.L.=\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(0+h)$
$\Rightarrow R.H.L.=\underset{h\to 0}{lim}f\left(h\right)$
$\Rightarrow R.H.L.=\underset{h\to 0}{lim}(nh+m)$
$\Rightarrow R.H.L.=n\left(0\right)+m=m$
So, $\underset{x\to 0}{lim}f\left(x\right)$ exists if $m=n$

Step $2:$ Solve $\underset{x\to 1}{lim}f\left(x\right)$
Limit exists at $x=1$ if
$L.H.L.=R.H.L.$

$L.H.L.=\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(1-h)$
$\Rightarrow L.H.L.=\underset{h\to 0}{lim}n(1-h)+m$
$\Rightarrow L.H.L.=n(1-0)+m$
$L.H.L.=n+m$

$R.H.L.=\underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(1+h)$
$\Rightarrow R.H.L.=\underset{h\to 0}{lim}\left(n\right(1+h{)}^3+m)$
$\Rightarrow R.H.L.=n(1+0{)}^3+m$
$\Rightarrow R.H.L.=n+m$
Since $L.H.L.=R.H.L.$
$m+n=m+n$
But this is always true
So $\underset{x\to 1}{lim}f\left(x\right)$ exists at all integral value of $m$ and $n$.