We have f (x) = $\begin{cases} mx^2 + n, & x < 0 \\ nx + m, & 0 \le x \le 1 \\ nx^3 + m, & x > 1 \end{cases}$ Now $\lim\limits_{x\to 0^{-}}$ f (x) = $\lim\limits_{x\to 0^{-}}$ $(mx^2+n)$ = m (0) + n = n and $\lim\limits_{x\to 0^{+}}$ f (x) = $\lim\limits_{x\to 0^{+}}$ (nx + m) = n (0) + m = m But for $\lim\limits_{x\to 0}$ f (x) to exist, we must have $\lim\limits_{x\to 0^{-}}$ f (x) = $\lim\limits_{x\to 0^{+}}$ f (x) i.e., n = m Hence $\lim\limits_{x\to 0}$ f (x) exists only if n = m. Now, $\lim\limits_{x\to 1^{-}}$ f (x) = $\lim\limits_{x\to 1^{-}}$ (nx + m) = n + m and $\lim\limits_{x\to 1^{+}}$ f (x) = $\lim\limits_{x\to 1^{+}}$ $(nx^3+m)$ = n + m ∴ Above condition shows that $\lim\limits_{x\to 1^{-}}$ f (x) = $\lim\limits_{x\to 1^{+}}$ f (x) = m + n. Thus $\lim\limits_{x\to 1}$ f (x) exists for all n , m.