Limit Evaluation for (1-tan x) / (sin x - cos x) as x approaches pi/4

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Homework Statement



lim (1-tan x) / (sin x - cos x)
x-->pi/4

Homework Equations





The Attempt at a Solution



Im not sure where to start with this one. Should I turn tan x into sinx/cos x and try to do cancel some of the equation out?
 
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That might work. Another approach is L'Hopital's Rule if you know it, since this expression has the indeterminate form 0/0.
 
meeklobraca said:

Homework Statement





Im not sure where to start with this one. Should I turn tan x into sinx/cos x and try to do cancel some of the equation out?

I just tried what you suggest and it came out okay.
 
Well I did some work with the equation and got 2/sqrt 2

This is how I got there,

I made the 1 = cos x/cos x
The tan x = sin x/cos x

Then I brought the denomenator up which canceled out both sin x - cos x which left me with 1 / -cos x

I plugged in the pi/4 which is sqrt 2/2 , which equals -2/sqrt2, which equals -sqrt2

What do you guys think?
 
Last edited:

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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