Question

If f(x) = f(x)=\left\\{\begin{matrix} mx^2+n, &\,\,\,\,x<0 \\\ nx+m,&\,\,\,\, 0\leq x\leq1 \\\ nx^3+m,&\,\,\,\, x>1 \end{matrix}\right.. For what integers m and n does both $\lim_{x\rightarrow 1}$f(x) and $\lim_{x\rightarrow 1}$ f(x) exist?

Answer 1

The given function is f(x)=\left\\{\begin{matrix} mx^2+n, &\,\,\,\,x<0 \\\ nx+m,&\,\,\,\, 0\leq x\leq1 \\\ nx^3+m,&\,\,\,\, x>1 \end{matrix}\right.
$\lim_{x\rightarrow 0^-}$f(x)=$\lim_{x\rightarrow 0^-}$(mx2+n)
=m(0)2+n
=n
$\lim_{x\rightarrow 0^+}$ f(x)= $\lim_{x\rightarrow 0^+}$(nx+m)
= n(0)+m
= m.
Thus, $\lim_{x\rightarrow 0}$f(x) exists if m = n.
$\lim_{x\rightarrow 0^-}$ f(x)= $\lim_{x\rightarrow 1^-}$(nx+m)
= n(1) +m
=m+n
$\lim_{x\rightarrow 1^+}$ f(x)= $\lim_{x\rightarrow 1^+}$(nx3+m)
= n(1) +m
=m+n
∴ $\lim_{x\rightarrow 1^-}$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.