# Thread: evaluating a trig limit

1. ## evaluating a trig limit

I am trying to evaluate lim of f(x)=COTx as x approaches $\frac{pi}{4}$ using the formula f'(a)= $\frac{cotx-cota}{x- a}$ and a= $\frac{pi}{4}}$
so..
$\frac{cotx-1}{x- \frac{pi}{4}}$

The answer is is suppose to be -2. How can I find the steps to get to -2?
I am sure I need to somehow use the trig limits $\frac{sinx}{x}=1$ and $\frac{cosx-1}{x}=0$

I start off like this, i suppose
$\frac{\frac{cos}{sin}-1}{x- \frac{pi}{4}}$

2. ## Re: evaluating a trig limit

Certainly you mean something other than the limit... you would not need that formula, and that is not what the limit is equal to:
lim cot&#40;x&#41; as x->pi&#47;4 - Wolfram|Alpha

Perhaps you mean the derivative of cotx? In that case, you want

$\lim_{x\to\tfrac{\pi}{4}} \frac{\cot x - \cot \tfrac{\pi} 4} {x - \tfrac{\pi} 4}$

which is what you have near the end there... I am just clarifying.

Have you learned the derivative of sine and cosine yet?

3. ## Re: evaluating a trig limit

I believe the OP is trying to find $f'\left(\frac{\pi}{4}\right)$ using the limit process where $f(x) = \cot{x}$

4. ## Re: evaluating a trig limit

$f'(a) = \lim_{x \to a} \frac{\frac{\cos{x}}{\sin{x}} - \frac{\cos{a}}{\sin{a}}}{x-a}$

$f'(a) = \lim_{x \to a} \frac{\frac{\cos{x}\sin{a} - \sin{x}\cos{a}}{\sin{x}\sin{a}}}{x-a}$

difference identity for sine ...

$f'(a) = \lim_{x \to a} \frac{\sin(a-x)}{(x-a)\sin{x}\sin{a}}$

note that $\sin(a-x) = -\sin(x-a)$ since sine is an odd function...

$f'(a) = \lim_{x \to a} -\frac{\sin(x-a)}{(x-a)\sin{x}\sin{a}}$

$f'(a) = \lim_{x \to a} -\frac{\sin(x-a)}{x-a} \cdot \frac{1}{\sin{x}\sin{a}}$

$f'(a) = \lim_{x \to a} -1 \cdot \frac{1}{\sin{x}\sin{a}} = -\frac{1}{\sin^2{a}}$