Question

Solve $g\left(x\right)=\frac{1-\sqrt{x}}{({\cos }^{-1}x{)}^2}$ as $x\to 1$.

Answer 1

$g\left(x\right)=\frac{1-\sqrt{x}}{({\cos }^{-1}x{)}^2}$

as $x\to 1$, $1-\sqrt{1}=0$ and ${\cos }^{-1}1=0$

So, $g\left(x\right)$ for $x$ tending to $1$ is $\frac{0}{0}$ form

Apply L'Hospital rule i.e. differentiating both numerator and denominator

Also use $\frac{d}{dx}{\cos }^{-1}x=-\frac{1}{\sqrt{1-x^2}}$

$\Rightarrow \underset{x\to 1}{lim}\frac{1-\sqrt{x}}{({\cos }^{-1}x{)}^2}=\underset{x\to 1}{lim}\frac{\sqrt{(1-x^2)}}{4\sqrt{x}{\cos }^{-1}x}$ on applying L'Hospital rule

Again it reduces to $\frac{0}{0}$ form, so again apply L'Hospital rule

$\Rightarrow \underset{x\to 1}{lim}\frac{\sqrt{(1-x^2)}}{4\sqrt{x}{\cos }^{-1}x}=\underset{x\to 1}{lim}\frac{x\sqrt{x}\sqrt{1-x^2}}{\sqrt{1-x^2}(4x-2\sqrt{1-x^2}{\cos }^{-1}x)}$

Now putting limit $x=1$ gives $\frac{1}{4}$ as the answer