Question

The value of $\underset{x\to 1}{lim}-\frac{(co{s}^{-1}x{)}^2}{1-\sqrt{x}}$ is equal to 4

Answer 1

$L=\underset{x\to 1}{lim}-\frac{(co{s}^{-1}x{)}^2}{1-\sqrt{x}}\to ln\frac{0}{0}$ form.
Apply L’Hospital’s Rule
$L=\underset{x\to 1}{lim}\frac{2co{s}^{-1}x\left(\frac{-1}{\sqrt{1-x^2}}\right)}{0-\frac{1}{2x^{\frac{1}{2}}}}L=\underset{x\to 1}{lim}\frac{4co{s}^{-1}\left(x\right).\left(x^{\frac{1}{2}}\right)}{\sqrt{1-x^2}}Letx=cos\theta whenx=5,\theta \to 0\sqrt{1-x^2}=sin\theta \therefore L=\underset{\theta \to 0}{lim}\frac{4co{s}^{-1}\left(cos\theta \right)(cos\theta {)}^{\frac{1}{2}}}{sin\theta }\therefore L=4\underset{x\to 1}{lim}\frac{\theta \sqrt{cos\theta }}{sin\theta }$
Divide N and D by $\theta$,
$L=\frac{\underset{\theta \to 0}{lim}\sqrt{cos\theta }}{\underset{\theta \to 0}{lim}\frac{sin\theta }{\theta }}\therefore L=\frac{4\times 1}{1}\therefore L=4$.