the answer is 0 but I don't know how to get itFind the derivative of ln e^u/u
lim
h-> 0 square root of (25-h) -5/h
The answer is 1/10 but I once again don't know how to do it.
If I'm reading it right it's $\displaystyle \lim_{h \rightarrow 0} \frac{\sqrt{25-h}-5}{h}$
Multiply the numerator and denominator by $\displaystyle \sqrt{25-h}+5$
$\displaystyle \frac{\sqrt{25-h}-5}{h} * \frac{\sqrt{25-h}+5}{\sqrt{25-h}+5}$
This comes out to be $\displaystyle \frac{-h}{h(\sqrt{25-h}+5)}=\frac{-1}{\sqrt{25-h}+5}$
Now try the limit again.