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

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Let f(x0 = (tanx -1)/(sin^2 x - cos^2 x)

lim (tanx - 1) /[ sin^2 x - cos^2 x)     x --> pi/4

First let us substitute with x = pi/4

==> lim f(x) = ( 1-1)/(1/2 - 1/2) = 0/0 ( the method failed)

But we know that: tanx = sinx/cosx

==< lim f(x) = lim (sinx/cosx -1)/(sin^2 x - cos^2 x)

                  = lim (sinx - cosx)/cosx(sin^2 x - cos^2 x)

                   = lim (sinx - cosx)/ cosx(sin x-cosx )(sinx + cosx)

                    = lim 1/cosx(sinx+cosx)

 ==> lim f(x) x---> pi/4 = 1/cospi/4(cospi/4 + sinpi/4)

                                    = 1/(sqrt2/2)*(sqrt2)

                                    = 2/2= 1

Then lim f(x) when x--> pi/4 =  1

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