Question

Show that the function $f\left(x\right)$ defined by
$f\left(x\right)=\frac{\sin x}{x}+\cos xx>0$
$=2x=0$
$=\frac{4\left(1-\sqrt{1-x}\right)}{x}x<0$ is continuous at $x=0$

Answer 1

Solution-

$f\left(x\right)=\{\begin{array}{cc}\frac{\sin x}{x}+\cos x& x>0\\ 2& x=0\\ \frac{4(1-\sqrt{1-x})}{x}& x<0\end{array}$

$f\left(0\right)=2$

LHL

$f\left(0^{-}\right)=\underset{x\to 0}{lim}\frac{4(1-\sqrt{1-x})}{x}$

$=\frac{-4}{-2\sqrt{1-x}}x\to 0^{-}$ [ Apply L-Hospital]

$=2$

RHL

$f\left(0^{+}\right)=\underset{x\to 0^{+}}{lim}\frac{\sin x}{x}+\cos x$

We know that $\underset{x\to 0^{+}}{lim}\frac{\sin x}{x}=1$

$f\left(0^{+}\right)=1+1=2$.

$f\left(0^{-}\right)=f\left(0\right)=f\left(0^{+}\right)$

Fraction is continuous at $x=0$.