Limit of another trigonometric function

[tmt1]

\(\displaystyle \lim_{{x}\to{\pi/4}} \frac{1-\tan(x)}{\sin(x)-\cos(x)}\)

So using, L'Hospital's rule, I get:

\(\displaystyle \lim_{{x}\to{\pi/4}} \frac{\sec^2(x)}{\cos(x)+\sin(x)}\)

But $\cos(x)+\sin(x) = 0$ when $x = \dfrac{\pi}{4}$ which is an indeterminate form, so how do I go from here?

 

[MarkFL]

\(\displaystyle \lim_{{x}\to{\pi/4}} \frac{1-\tan(x)}{\sin(x)-\cos(x)}\)

So using, L'Hospital's rule, I get:

\(\displaystyle \lim_{{x}\to{\pi/4}} \frac{\sec^2(x)}{\cos(x)+\sin(x)}\)

But $\cos(x)+\sin(x) = 0$ when $x = \dfrac{\pi}{4}$ which is an indeterminate form, so how do I go from here?

Using L'Hôpital's rule, you should get:

\(\displaystyle \lim_{{x}\to{\pi/4}} \frac{-\sec^2(x)}{\cos(x)+\sin(x)}\)

Rethink:

\(\displaystyle \cos\left(\frac{\pi}{4}\right)+\sin\left(\frac{\pi}{4}\right)\)...:D

 

[tmt1]

Using L'Hôpital's rule, you should get:

\(\displaystyle \lim_{{x}\to{\pi/4}} \frac{-\sec^2(x)}{\cos(x)+\sin(x)}\)

Rethink:

\(\displaystyle \cos\left(\frac{\pi}{4}\right)+\sin\left(\frac{\pi}{4}\right)\)...:D

Thanks,

I get

$-\frac{2}{\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}}$

which is equal to

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

.

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

$- \frac{2}{\sqrt{2}}$

But the answer is just $-\sqrt{2}$

 

[MarkFL]

\(\displaystyle -\frac{2}{\sqrt{2}}=-\frac{\sqrt{2}\cdot\cancel{\sqrt{2}}}{\cancel{\sqrt{2}}}=-\sqrt{2}\)

You had the correct answer, it could just be further simplified. :D