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.