Question

Discuss the continuity of function $f$ at $x=0$
Where $f\left(x\right)=\frac{\sqrt{4+x}-2}{3x}$, for $x\ne 0$
$=\frac{1}{12}$, for $x=0$

Answer 1

$f\left(x\right)=\frac{\sqrt{4+x}-2}{3x}$ for $(x\ne 0)$

& $f\left(0\right)=\frac{1}{12}$

$\therefore$If $f\left(0\right)={lim}_{x\to 0}f\left(x\right)$ then function is continuous

$x=0$

${lim}_{x\to 0}f\left(x\right)={lim}_{x\to 0}\frac{\sqrt{4+x}-2}{3x}$

As it is of $\frac{0}{0}$ form, we use L'Hospital's rule

$\therefore$ differentiating numerator and denominator

$\therefore {lim}_{x\to 0}f\left(x\right)={lim}_{x\to 0}\frac{\frac{1}{2\sqrt{4+x}}}{3}=\left(\frac{1}{2\sqrt{4}}\right)\times \frac{1}{3}$

$=\frac{1}{2\times 2}\times \frac{1}{3}=\frac{1}{12}$

$\therefore f\left(0\right)=\frac{1}{12}$ & ${lim}_{x\to 0}f\left(x\right)=\frac{1}{12}$

Limit exists, function is continuous