Question

Examine the continually of $f\left(x\right)=\left|x\right|+|x-1|$ in the interval (1, -2) Find the value of f(0) if the following function is continous at x=0 $f\left(x\right)=\frac{\sqrt{1+x}-\sqrt[3]{31+x}}{x}$

Answer 1

Function is continous at $x=0$

$\therefore \underset{h\to 0}{lim}f\left(0\right)$

$\Rightarrow \underset{h\to 0}{lim}\frac{\sqrt{1+x}-\sqrt[3]{31+x}}{x}=f\left(0\right)$

$\Rightarrow \underset{h\to 0}{lim}\frac{(1+x{)}^{1/2}-(1+x{)}^{1/3}}{x}=f\left(0\right)$

$\Rightarrow \underset{h\to 0}{lim}\frac{(1+\frac{1}{2}x+...)-(1+\frac{1}{3}x+..)}{x}=f\left(0\right)$

$\Rightarrow \underset{h\to 0}{lim}\frac{(\frac{1}{2}-\frac{1}{3})x+...xuptohigherpowers}{x}=f\left(0\right)$

$\Rightarrow \underset{h\to 0}{lim}\frac{x[\frac{1}{6}+...xuptohigherpowers]}{x}=f\left(0\right)$

$\Rightarrow \frac{1}{6}=f\left(0\right)$

Hence, $x=\frac{1}{6}$