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If $f(x)=\left\{\begin{array}{l}m x^2+n, x < 0 \\ n x+m, 0 \leq x \leq 1, \text { For what integers } m \text { and } \\ n x^3+m, x>1\end{array}\right.$ $n$ does both $\lim _{x \rightarrow 0} f(x)$ and $\lim _{x \rightarrow 1} f(x)$ exist?
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(i) When $x < 0, f(x)=m x^2+n$
$\therefore \quad \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}\left(m x^2+n\right)=n$ when $x>0, f(x)=n x+m$
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}}(n x+m)=m$
$\therefore \quad$ for $\lim _{x \rightarrow 0} f(x)$ to exists, we need $n=m$
(ii) When $x < 1, f(x)=n x+m$;
$\begin{array}{ll}
& \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}(n x+m)=m+n \\
& \text { when } x>1, f(x)=n x^3+m \\
\therefore \quad & \lim _{x \rightarrow 1^{+}} f(x)=\left(n x^3+m\right) \\
& =\lim _{x \rightarrow 1^{+}}\left(n x^3+m\right)=n+m \\
& \text { When } m=n, \forall m \in R \\
\therefore \quad & \lim _{x \rightarrow 0} f(x)=m \text { and } \\
& \lim _{x \rightarrow 1} f(x)=2 m, \forall m \in R
\end{array}$
$\therefore \quad \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}\left(m x^2+n\right)=n$ when $x>0, f(x)=n x+m$
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}}(n x+m)=m$
$\therefore \quad$ for $\lim _{x \rightarrow 0} f(x)$ to exists, we need $n=m$
(ii) When $x < 1, f(x)=n x+m$;
$\begin{array}{ll}
& \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}(n x+m)=m+n \\
& \text { when } x>1, f(x)=n x^3+m \\
\therefore \quad & \lim _{x \rightarrow 1^{+}} f(x)=\left(n x^3+m\right) \\
& =\lim _{x \rightarrow 1^{+}}\left(n x^3+m\right)=n+m \\
& \text { When } m=n, \forall m \in R \\
\therefore \quad & \lim _{x \rightarrow 0} f(x)=m \text { and } \\
& \lim _{x \rightarrow 1} f(x)=2 m, \forall m \in R
\end{array}$
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