Question

Take the limit of this trig function:

Answer 1

\[\lim_{x \rightarrow \pi/4}\frac{ 7-7tanx }{ sinx-cosx }\]

Answer 2

use L'Hopital's rule:
\[\frac{-7\sec^2(x)}{\cos(x)+\sin(x)}\]

Answer 3

take the limit as x goes to pi/4:
\[\frac{-7\sec^2(\pi/4)}{\cos(\pi/4)+\sin(\pi/4)} = \frac{-7(2)}{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}}\]

Answer 4

-7/sqrt(2)?

Answer 5

No. That can't be right. I must have made a mistake in my algebra.

Answer 6

I'm lost. -7/sqrt(2) is not the right answer.

Answer 7

I didn't use l'hopital's rule because we haven't gotten there yet.

Answer 8

I multiplied top and bottom by cosx

Answer 9

and pulled out the 7

Answer 10

so I was left with \[\frac{ cosx-sinx }{ sinx-cosx }\]

Answer 11

sorry, that's wrong

Answer 12

should be a cosx multiplied in the denominator

Answer 13

\[\frac{ cosx-sinx }{ cosx(sinx-cosx) }\]

Answer 14

\[\frac{2}{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}} = \sqrt{2}\]

Answer 15

oh, so it's just \[7\sqrt{2}\]

Answer 16

the answer is:
\[7\sqrt{2}\]

Answer 17

?

Answer 18

yes

Answer 19

Hmm, it's telling me that the answer is incorrect.

Answer 20

Hmm this is what i'm getting