Question

Let $f\left(x\right)=\{\begin{array}{cc}\sin x,& x\le 0\\ x^2+l,& 0<x<1\\ mx+3,& 1\le x\le 3\end{array}$. If both $\underset{x\to 0}{lim}f\left(x\right)$ and $\underset{x\to 1}{lim}f\left(x\right)$ exist, then the value of $\underset{x\to 5}{lim}(l{x}^2+mx+17)$ is equal to

Answer 1

$f\left(x\right)=\{\begin{array}{cc}\sin x,& x\le 0\\ x^2+l,& 0<x<1\\ mx+3,& 1\le x\le 3\end{array}$

$\underset{x\to 0}{lim}f\left(x\right)$ exists
$\Rightarrow \underset{x\to 0^{-}}{lim}\left(\sin x\right)=\underset{x\to 0^{+}}{lim}(x^2+l)\Rightarrow 0=l\cdots \left(1\right)$

$\underset{x\to 1}{lim}f\left(x\right)$ exists
$\Rightarrow \underset{x\to 1^{-}}{lim}(x^2+l)=\underset{x\to 1^{+}}{lim}(mx+3)\Rightarrow 1+l=m+3\cdots \left(2\right)$

From $\left(1\right)$ and $\left(2\right)$
$\Rightarrow l=0,m=-2$
$\therefore \underset{x\to 5}{lim}(l{x}^2+mx+17)=\underset{x\to 5}{lim}(17-2x)=7$