Finding a limit using substitution rule, my answer is 0, my book`s is -2

[wajed]

Homework Statement

Lim [(tanx)^2] / [1 + secx] <<< as x goes to pi

Homework Equations

The Attempt at a Solution

(tan x)^2 = (sin x)^2 / (cos x)^2

(sin x)^2 = y

lim y = 0 <<< as x goes to pi


lim [y/ (cos x)^2] / [1 + (1/y)] <<< as y goes to C=0

1+ (1/y) = (y+1)/y

lim [y^2] / (y+1) (cos x)^2

y=0

so, 0/(0+1)(cos 0)^2 = 0/1(1) = 0/1 = 0

why does my book mentions that answer is -2?

 

[wajed]

huh, that was a stupid mistake, sorry.
(I`m just totally nervous, I got an exam tomorrow)EDIT:
but, how do I solve that anyway?

I can`t manage to do it..

I know the whole thing is about the "cos x" and that I have to make it turn to something in terms of y, but how?

 

[slider142]

sec(x) is not 1/sin(x). It's 1/cos(x). Try the substitution u = cos(x).

 

[quantumdude]

lim [y/ (cos x)^2] / [1 + (1/y)]

Not quite. If $\sin^2(x)=y$ then $\cos^2(x)=1-y$. Here's the tricky part: $\sec(x)=1/\sqrt{1-y}$ if $\cos(x)\geq0$ but $\sec(x)=-1/\sqrt{1-y}$ if $\cos(x)<0$.