Question

(1-tan^2 x)/(cos x-sin x) x to pi/4

Answer 1

plsss help

Answer 2

OK lets do this :)
I believe you are looking for the limit at pi/4

First step is to check what values it takes at pi/4
tan(pi/4)=1, thus 1-tan^2(pi/4)=0
Cos(pi/4)-sin(pi/4)=0

Answer 3

Now we have a limit with 0/0, that can be anything. We do not know what it will take.
However we can use the L`Hopital rule. Do you know what that is?

Answer 4

if you want to do it without L'Hopital's,
write tan^2x = sin^2x/cos^2x

Answer 5

If you follow hartnn`s idea than first lets just deal with the numerator:
1-tan^2x=1-sin^2x/cos^2x
(1=cos^2x/cos^2x)
thus
1-sin^2x/cos^2x=(cos^2x-sin^2x)/cos^2x

cos^2x-sin^2x=(cosx-sinx)(cosx+sinx)

So neatly written (the full expression is)
\[\frac{ \frac{ {(cosx -sinx)(cosx+sinx)} }{ \cos^2x }}{cosx -sinx}\]

Answer 6

You can cancel out cosx-sinx to get

Answer 7

\[\frac{ {cosx+sinx} }{\cos^2x }\]

Answer 8

And this is not 0/0 at pi/4.
The answer is \[2\sqrt{2}\]

Answer 9

2 sqrt 2 or just sqrt 2 ?

Answer 10

\[2\sqrt{2}\]
What did you get?

Answer 11

yup, 2 sqrt 2 is correct

Answer 12

thanks a lot people no i dont know l hospitals rule thanks for the alternate method