Question

Can L' Hospital's rule the applied to evaluate the limit
lim x tends to (pi)/2 (1-sin x)/cos x
if yes , evaluate the limit .

Answer 1

Answer on Question # 41154 – Math - Integral Calculus

Can L' Hospital's rule be applied to evaluate the limit lim x tends to (pi)/2 (1-sin x)/cos x. If yes, evaluate the limit.

Solution.

Because the functions $f(x) = 1 - \sin x$ and $g(x) = \cos x$ are differentiable on an open interval containing pi/2 and $f\left(\frac{\pi}{2}\right) = g\left(\frac{\pi}{2}\right) = \left[\frac{0}{0}\right], g'(x) \neq 0$ we can apply L'Hospital's rule.

We have the limit:

$$\lim_{x \to \frac{\pi}{2}} \frac{1 - \sin x}{\cos x}$$

Evaluate the limit:

$$\lim_{x \to \frac{\pi}{2}} \frac{1 - \sin x}{\cos x} = \left[ \frac{1 - \sin \frac{\pi}{2}}{\cos \frac{\pi}{2}} \right] = \left[ \frac{0}{0} \right]$$

Then we can use L'Hospital's rule:

$$\lim_{x \to \frac{\pi}{2}} \frac{1 - \sin x}{\cos x} = \lim_{x \to \frac{\pi}{2}} \frac{-\cos x}{-\sin x} = \lim_{x \to \frac{\pi}{2}} \cot x = \cot \frac{\pi}{2} = 0$$

Answer:

$$\lim_{x \to \frac{\pi}{2}} \frac{1 - \sin x}{\cos x} = 0$$