Question

Using L' Hopital's rule, find the limit of

Answer 1

Answer:

L=0

Step-by-step explanation:

$L=\lim\limits_{x \rightarrow \frac{\pi}{2}}3secx-3tanx$

Replacing the value of x we get ∞ - ∞ which is an indetermined expression

We must transform the limit so it can be shown as a fraction and the L'Hopital's rule can be applied:

$L=\lim\limits_{x \rightarrow \frac{\pi}{2}}\frac{3-3sinx}{cosx}=\frac{0}{0}$

Now we can take the derivative in both parts of the fraction

$L=\lim\limits_{x \rightarrow \frac{\pi}{2}}\frac{-3cosx}{-sinx}=3\lim\limits_{x \rightarrow \frac{\pi}{2}}\frac{cosx}{sinx}=3\times 0=0$