lim(tanx-1)/[(sinx)^2 - (cosx)^2] x->pi/4

First, we'll verify if it is a case of indeterminacy, so
we'll substitute x by pi/4, into the given
expression:

f(pi/4) = (tan pi/4-1)/[(sin pi/4)^2 - (cos
pi/4)^2]  , where tan pi/4= 1, cos pi/4  =sin pi/4 
=sqrt2/2

f(pi/4) = (1-1)/(1/2 -
1/2)

f(pi/4) = 0/0,  indeterminacy
case

We'll solve the limit, by substituting the
difference

tan x - 1 = (sin x / cos x) - 1 = (sin x - cos
x)/ cos x

lim(tanx-1)/[(sinx)^2 - (cosx)^2] = lim (sin x -
cos x)/ cos x* [(sinx)^2 - (cosx)^2]

We'll write the
difference of square from denominator as:

[(sinx)^2 -
(cosx)^2] = (sinx-cosx)(sinx+cosx)

lim (sin x-cos x)/ cos
x*[(sinx)^2-(cosx)^2]=lim (sin x-cos
x)/cosx*(sinx-cosx)(sinx+cosx)

We'll reduce like terms,(sin
x-cos x):

lim 1/cosx*(sinx+cosx) =1/cos pi/4*(sin pi/4 +
cos pi/4)

lim 1/cosx*(sinx+cosx) =
2/(sqrt2)(sqrt2)

lim 1/cosx*(sinx+cosx) =
2/2

lim 1/cosx*(sinx+cosx) =
1