Trig limit help?

Question

limit as x approaches pi over 4

(6-6tanx)/(sinx-cosx)

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Answer 1

Substitute t=x-Pi/4, i.e. x=t+Pi/4
tanx = (1 + tant)/(1 - tant) = (cost + sint)/(cost - sint)
sinx - cosx = sqrt(2) sint
(6-6tanx)/(sinx-cosx) = -12sint/[(cost-sint)sqrt(2)sin(t)] = -6sqrt(2)/(cost-sint)
At t->0, you get limit -6sqrt(2)

Check my calculations.

Sorry, there is a simpler way
1 - tanx = (cosx - sinx)cosx, so
(6-6tanx)/(sinx-cosx)=-6/cosx
So lim at x->pi/4 is -6sqrt(2)

Answer 2

Since the limit is 0/0, which is indeterminate in its initial form, you can use L'Hospital's rule on this limit to make it work.

Derivative of the top: -6(sec(x))^2
Derivative of the bottom: cos(x) + sin(x)

So, it is not the limit as x approaches pi/4:

-6(sec(x))^2/(cos(x) + sin(x))

Which would be -12/sqrt(2)

=

-6*sqrt(2)