# Math Help - Using limits to say where a derivative is defined

1. ## Using limits to say where a derivative is defined

I have been asked to say where the following derivative is defined;

f'(x)=cosx-sinx for 2kPi-Pi/2<x<2kPi+Pi/2
=cosx+sinx for 2kPi+Pi/2<x<2kPi+3Pi/2

Im pretty sure i need to be using limits as I hav seen from a similar question but I am unsure how?

Thanks for any help

2. Hello, leigh!

Where is the following derivative defined?

. . $f'(x)\;=\; \begin{Bmatrix}\cos x-\sin x & \text{for }2k\pi-\frac{\pi}{2} < x < 2k\pi+\frac{\pi}{2} \\ \\[-3mm]
\cos x+ \sin x & \text{for }2k\pi+\frac{\pi}{2}

For clarity, I wrote the function like this:

$f'(x) \;=\;\begin{Bmatrix}\cos x - \sin x & x \in (\text{-}\frac{\pi}{2}.\,\frac{\pi}{2}) & (\frac{3\pi}{2},\,\frac{5\pi}{2}) & (\frac{7\pi}{2},\,\frac{9\pi}{2}) & \hdots \\ \\[-3mm] \cos x + \sin x & x \in (\frac{\pi}{2},\,\frac{3\pi}{2}) & (\frac{5\pi}{2},\,\frac{7\pi}{2}) & (\frac{9\pi}{2},\,\frac{11\pi}{2}) & \hdots \end{array}$

Clearly the derivative is defined within those intervals.
The only doubts occur at the endpoints.

Test an endpoint using one-sided limits.

. . $\displaystyle \lim_{x\to\frac{\pi}{2}^-} f'(x) \;=\;\lim_{x\to\frac{\pi}{2}^-}(\cos x - \sin x) \;=\;\cos\tfrac{\pi}{2} - \sin\tfrac{\pi}{2} \;=\;0 - 1 \;=\;-1$

. . $\displaystyle \lim_{x\to\frac{\pi}{2}^+}f'(x) \;=\;\lim_{x\to\frac{\pi}{2}^+}(\cos x + \sin x) \;=\;\cos\tfrac{\pi}{2} + \sin\tfrac{\pi}{2} \;=\;0 + 1 \;=\;+1$

The limit does not exist at $x = \frac{\pi}{2}.$

Hence, the derivative is not defined at $x = \frac{\pi}{2}.$

It can be shown that the derivative is undefined for any odd multiple of $\frac{\pi}{2}.$

$f'(x)$ is defined for all $\,x$ except at $x \:=\:\dfrac{(2k+1)\pi}{2}$ for any integer $\,k.$

3. The limit does not exist at Pi/2
How do you know this from finding the limit?

Thank you very much for your help

4. o