Question

The value of f(0) so that the function
$f\left(x\right)=\frac{\sqrt{1+x}-\sqrt[3]{31+x}}{x}$
becomes continuous, is equal to

• A. $\frac{1}{6}$
• B. $\frac{1}{4}$
• C. 2
• D. $\frac{1}{3}$

Answer 1

The correct option is A $\frac{1}{6}$

The function $f\left(x\right)$ is continuous except possibly at $x=0$. For $f$ to be continuous at $x=0$, we must have
$f\left(0\right)=\underset{x\to 0}{lim}f\left(x\right)=\underset{x\to 0}{lim}\frac{(1+x{)}^{1/2}-(1+x{)}^{1/3}}{x}$
$=\underset{x\to 0}{lim}\frac{[1+\frac{1}{2}x+\frac{\left(1/2\right)(-1/2)}{2}{x}^2+...]-[1+\frac{1}{3}x+\frac{\left(1/3\right)(-2/3)}{2}{x}^2+...]}{x}$
Therefore,
$f\left(0\right)=\underset{x\to 0}{lim}x\frac{[\frac{1}{2}+\frac{\left(1/2\right)(-1/2)}{2}x+...]-[\frac{1}{3}+\frac{\left(1/3\right)(-2/3)}{2}x+...]}{x}$

$=\underset{x\to 0}{lim}[(\frac{1}{2}-\frac{1}{3})+termcontainingx]=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$
(Alternatively apply L-Hospital's Rule)