Question

lim x-> pi/4 of (1-tanx)/(sinx-cosx)

Answer 1

Did you try multiplying by conjugates?

Answer 2

No I multiplied by cosx?

Answer 3

Why, though?

Answer 4

Because tanx becomes sinx/cosx so to cancel the denominator I multiplied by cos x so it becomes (cosx-sinx)/cosx•sinx-cosx and then the cosines equal 1 and as well do the sines so its -1/-cosx

Answer 5

But the answer is supposed to be -square root of 2 but I got positive and don't know what i did wrong

Answer 6

Fair enough, hang on.

Answer 7

\[\Large \lim_{x\rightarrow \frac \pi 4}\frac{1-\tan(x)}{\sin(x) - \cos(x)}\]

Answer 8

You missed a tiny detail, it would seem ^_^
\[\Large \frac{\cos(x)}{\cos(x)}\cdot \frac{1-\tan(x)}{\sin(x) - \cos(x)}= \frac{\cos(x)-\sin(x)}{\cos(x)[\sin(x) - \cos(x)]}\]

Answer 9

See it now?
\[\LARGE = \frac{\cos(x) - \sin(x)}{\cos(x)[\cos(x) -\sin(x)]\color{red}{(-1)}}\]

Answer 10

Welcome to OpenStudy! Message me if you have any questions!

Answer 11

Because these two are negatives of each other:
\[\LARGE\frac{\color{red}{\cos(x)-\sin(x)}}{\cos(x)[\color{blue}{\sin(x) - \cos(x)}]}\]

Answer 12

I'm new to this website but for some reason I can't see all of what you wrote

Answer 13

Where you put you missed a tiny detail I can't see the whole problem and where you put see it now as well

Answer 14

@Daisytoolovely sorry, was distracted...
basically
cos(x) - sin(x) = -[sin(x) - cos(x)]
So that in effect,
cos(x) - sin(x) / sin(x) - cos(x) = -1