Question
If $\text{f}(\text{x})=\begin{cases}a & x = 0\\b & x > 0\end{cases}\begin{cases}\text{mx}^2+\text{n,} & x < 0\\\text{nx}+m, & 0\leq\text{x}\leq 1\\\text{nx}^3+\text{x}&\text{x}>1\end{cases}$ . For what integers m and n does both $\lim\limits_{\text{x}\rightarrow0}\text{f}(\text{x})$ and $\lim\limits_{\text{x}\rightarrow1}\text{f}(\text{x})$exist?

Answer

The given function is$\text{f}(\text{x})=\begin{cases}a & x = 0\\b & x > 0\end{cases}\begin{cases}\text{mx}^2+\text{n,} & x < 0\\\text{nx}+m, & 0\leq\text{x}\leq 1\\\text{nx}^3+\text{x}&\text{x}>1\end{cases}$
$\lim\limits_{\text{x}\rightarrow0^-}\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow0}(\text{mx}^2+\text{n})$
$=\text{m}(0)^2+\text{n}$
$=\text{n}$ $\lim\limits_{\text{x}\rightarrow0^+}=\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow0}(\text{nx}+\text{m})$ $=\text{n}(0)+\text{m}$ $=\text{m}$. Thus, $\lim\limits_{\text{x}\rightarrow0}\text{f}(\text{x})$exists if m = n . $\lim\limits_{\text{x}\rightarrow1^-}\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow1}(\text{nx}+\text{m})$ $=\text{n}(1)+\text{m}$$=\text{m+n}$
$\lim\limits_{\text{x}\rightarrow1^+}\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow1}(\text{nx}^3+\text{m})$ $=\text{n}(1)^3+\text{m}$ $=\text{m+n}$ $\therefore\lim\limits_{\text{x}\rightarrow1^-}\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow1^+}\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow1}\text{f}(\text{x})$. Thus, $\lim\limits_{\text{x}\rightarrow1}\text{f}(\text{x})$exists for any integral value of m and n.

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