For what integers m and n does

Question:

If $f(x)= \begin{cases}m x^{2}+n, & x<0 \\ n x+m, & 0 \leq x \leq 1 . \text { For what integers } m \text { and } n \text { does } \lim _{x \rightarrow 0} f(x) \text { and } \lim _{x \rightarrow 1} f(x) \text { exist? } \\ n x^{3}+m, & x>1\end{cases}$

Solution:

The given function is

$f(x)= \begin{cases}m x^{2}+n, & x<0 \\ n x+m, & 0 \leq x \leq 1 \\ n x^{3}+m, & x>1\end{cases}$

$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0}\left(m x^{2}+n\right)$

$=m(0)^{2}+n$

$=n$

$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0}(n x+m)$

$=n(0)+m$

$=m .$

Thus, $\lim _{x \rightarrow 0} f(x)$ exists if $m=n$.

$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1}(n x+m)$

$=n(1)+m$

$=m+n$

$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1}\left(n x^{3}+m\right)$

$=n(1)^{3}+m$

$=m+n$

$\therefore \lim _{x \rightarrow 1} f(x)=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1} f(x) .$

Thus, $\lim _{x \rightarrow 1} f(x)$ exists for any integral value of $m$ and $n$.

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