Question

If f(x) = $\{\begin{array}{cc}m{x}^2+n,& x<0\\ nx+m,& 0\le x\le 1\\ n{x}^3+m,& x>1\end{array}$ for what intergers m and n does both $\underset{x\to 0}{\text{lim}}$ f(x) and $\underset{x\to 1}{\text{lim}}$ f(x) exist?

Answer 1

It is given that $\underset{x\to 0}{\text{lim}}$ f(x) and $\underset{x\to 1}{\text{lim}}$ f(x) both exist.

$\Rightarrow \underset{x\to 0^{-}}{\text{lim}}f\left(x\right)=\underset{x\to 0^{+}}{\text{lim}}$ f(x)

and $\underset{x\to 1^{-}}{\text{lim}}$ f(x) =$\underset{x\to 1^{+}}{\text{lim}}$ f(x)

Now $\underset{x\to 0^{-}}{\text{lim}}$ f(x) = $\underset{x\to 0^{-}}{\text{lim}}(m{x}^2+n)$

= $\underset{h\to 0}{\text{lim}}\left[m\right(0-h{)}^2+n]$

= $\underset{h\to 0}{\text{lim}}[m{h}^2+n]$ = n

= $\underset{x\to 0^{+}}{\text{lim}}$ f(x) = $\underset{h\to 0^{+}}{\text{lim}}$ (nx+m)

= $\underset{h\to 0}{\text{lim}}$ [n(0+h)+m]

= $\underset{h\to 0}{\text{lim}}$ (nh+m) = m

Now = $\underset{x\to 0^{-}}{\text{lim}}$ f(x) = $\underset{h\to 0^{+}}{\text{lim}}$ f(x) $\Rightarrow$ n = m ...(i)

For = $\underset{x\to 0}{\text{lim}}$ f(x) to exist we need m = n

Also = $\underset{x\to 1^{-}}{\text{lim}}$ = $\underset{x\to 1-}{\text{lim}}$ (nx+m)

=$\underset{h\to 0}{\text{lim}}$ [n(1-h)+m]

= $\underset{h\to 0}{\text{lim}}$ (n-nh + m) = n + m

$\underset{x\to 1^{+}}{\text{lim}}$ f(x) = $\underset{x\to 1^{+}}{\text{lim}}(n{x}^3+m)$

= $\underset{h\to 0}{\text{lim}}\left[n\right(1+h{)}^3+m]$

= $\underset{h\to 0}{\text{lim}}\left[n\right(1+h^3+3h^2+3h)+m]$

= n+m

Now $\underset{x\to 1^{-}}{\text{lim}}$ f(x) = $\underset{x\to 1^{+}}{\text{lim}}$ f(x) $\Rightarrow$ n+m= n+m

Thus $\underset{x\to 0}{\text{lim}}$ f(x) exists for any integral value of m and n, such that they are equal.