Question
If $f(x) = \left\{ {\begin{array}{*{20}{c}} {m{x^2} + n,} \\ {nx + m,} \\ {n{x^3} + m,} \end{array}} \right.\begin{array}{*{20}{c}} {x < 0} \\ {0 \le x \le 1} \\ {x > 1} \end{array}$ .
For what integers m and n does both $\mathop {\lim }\limits_{x \to 0} f(x)$ and $\mathop {\lim }\limits_{x \to 1} f(x)$ exist?

Answer

It is given that
$\mathop {\lim }\limits_{x \to 0}$ f(x) and $\mathop {\lim }\limits_{x \to 1}$ f(x) both exist.
$\Rightarrow \;\mathop {\lim }\limits_{x \to {0^ - }}$ f(x) and = $\mathop {\lim }\limits_{x \to {0^ +}}$f(x) = $\mathop {\lim }\limits_{x \to {1^ + }}$ f(x)
Now $\mathop {\lim }\limits_{x \to {0^- }}$ f(x) = $\mathop {\lim }\limits_{x \to {0 }} (mx^2+ n) = n$
$\mathop {\lim }\limits_{x \to {0^ + }}$ f(x) = $\mathop {\lim }\limits_{x \to {0 }}$ (nx + m) = m
Now $\mathop {\lim }\limits_{x \to {0^ + }}$ f(x) = $\mathop {\lim }\limits_{x \to {0^ + }}$ f(x) $\Rightarrow$ n = m ...(i)
For $\mathop {\lim }\limits_{x \to 0}$ f(x) to exist we need m = n
Also, $\mathop {\lim }\limits_{x \to {1^ - }}$ f(x) = $\mathop {\lim }\limits_{x \to {1}} (nx + m) = n + m$
$\mathop {\lim }\limits_{x \to {1^ + }}$ f(x) = $\mathop {\lim }\limits_{x \to {1}} (nx^3 + m) = n + m$
Now $\mathop {\lim }\limits_{x \to {1^ - }}$ f(x) = $\mathop {\lim }\limits_{x \to {1^ + }}$ f(x) $\Rightarrow$ n + m = n + m
Thus $\mathop {\lim }\limits_{x \to 0}$ f(x) exists for any integral value of m and n.

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