# Question #4ca29

Feb 15, 2017

$= - \frac{1}{2}$

#### Explanation:

It looks like:

${\lim}_{x \to \frac{\pi}{4}} \frac{\tan x - 1}{\sin 4 x}$

We can use L'Hôpital's Rule because this is in indeterminate form, ie if we plug the x-value straight in, we get $\frac{\tan x - 1}{\sin 4 x} = \frac{1 - 1}{0} = \frac{0}{0}$.

So after one round of L'Hôpital, we are at:

$= {\lim}_{x \to \frac{\pi}{4}} \frac{{\sec}^{2} x}{4 \cos 4 x}$

$= {\lim}_{x \to \frac{\pi}{4}} \frac{1}{4 {\cos}^{2} x \cos 4 x}$

If we plug the x-value straight in again, we get:

$\frac{1}{4 {\cos}^{2} \left(\frac{\pi}{4}\right) \cos \pi} = \frac{1}{4 \left(\frac{1}{2}\right) \left(- 1\right)} = - \frac{1}{2}$

Because ${\cos}^{2} x$ and $\cos 4 x$ are continuous at $x = \frac{\pi}{4}$, we can rely on that result.

Of course, that's a complete black-box. The more transparent way is to use, perhaps, a Taylor Expansion; or, maybe some algebraic manipulation of the trig functions.