# Thread: trigonometric limits

1. ## trigonometric limits

I'm having trouble following along with steps in my notes. Maybe someone can chime in and point me in the correct direction.

The original problem is:

$\displaystyle \displaystyle\lim_{ x\to\frac{\pi}{4} } \frac{1 - tan x}{ cos x - sin x}$

I am stuck trying to get from this step

$\displaystyle \displaystyle\frac{\cos{x} - ( \frac{\sin^2{x} }{ cosx }) }{ cos^2x - sin^2x})$

To

$\displaystyle \frac{(cos^2x - sin^2x }{ cosx }) ( \frac{1 }{ cos^2x - sin^2x})$

Any help is appreciated!

2. Originally Posted by louis
I'm having trouble following along with steps in my notes. Maybe someone can chime in and point me in the correct direction.

The original problem is:

$\displaystyle \displaystyle\lim_{ x\to\frac{\pi}{4} } \frac{1 - tan x}{ cos x - sin x}$

I am stuck trying to get from this step

$\displaystyle \displaystyle\frac{\cos{x} - ( \frac{\sin^2{x} }{ cosx }) }{ cos^2x - sin^2x})$

To

$\displaystyle \frac{(cos^2x - sin^2x }{ cosx }) ( \frac{1 }{ cos^2x - sin^2x})$

Any help is appreciated!
Note that

\displaystyle \begin{aligned}\dfrac{\cos x-\frac{\sin^2x}{\cos x}}{\cos^2x-\sin^2x}&=\dfrac{\frac{\cos^2x}{\cos x}-\frac{\sin^2x}{\cos x}}{\cos^2x-\sin^2x}\\ &= \dfrac{\frac{\cos^2x-\sin^2x}{\cos x}}{\cos^2x-\sin^2x}\\ &=\dfrac{\cos^2x-\sin^2x}{\cos x}\cdot\dfrac{1}{\cos^2x-\sin^2x}\end{aligned}

Does this make sense?

3. that is the step i was missing! thank you so much for the quick reply!