How do I approach solving this indeterminate limit? $\lim_\limits{h\to 0}\frac{1}{h}\ln\left(\frac{2+h}{2}\right)$ Disclaimer: I am a middle aged adult learning Calculus.  This is not a student posting his homework assignment.  Thank humanity for this great forum!
$$\lim_\limits{h\to 0}\frac{1}{h}\ln\left(\frac{2+h}{2}\right)$$
1) Can't directly sub the h.  So, normally you reduce and cancel.  Can you point me in the right direction?  The directions say "manipulate the expression so L'Hopital's is used" I think L'hopital's is involved.  Just not sure how to deal with the $\frac{1}{h}$
$$\lim_\limits{x\to \infty}\frac{x^k}{e^x}$$
2) Also, any tips on the one above?  If $k$ is a positive integer, what is the limit above?
Thanks for any guidance.
 A: If you must use l'Hospital's rule, then we can use it to note that:
$$\lim_{h\to 0}\frac{\ln(1+h/2)}{h}=\lim_{h\to0}\frac{1}{2}\frac{1}{1+h/2}$$
Direct substitution then gives us the answer:
$$\lim_{h\to0}\frac{1}{h}\ln\left(\frac{2+h}{2}\right) = \frac{1}{2}$$

For the second limit:
$$\lim_{x\to \infty}\frac{x^{k}}{e^{x}}$$
We note that we can write the Maclaurin series for $e^{x}$:
$$e^{x} = \sum_{k=0}^{\infty}\frac{x^{k}}{k!}$$
And so we can write the limit:
$$\lim_{x\to\infty}\frac{x^{k}}{\sum_{i=0}^{\infty}\frac{x^{i}}{i!}}$$
What does this mean for any integer value of $k$?
A: $\ln(1+au)\sim_0 au$, hence $\dfrac1h\ln\Bigl(\dfrac{2+h}2\Bigr)=\dfrac1h\ln\Bigl(1+\dfrac h2\Bigr)\sim_0 \dfrac{\dfrac h2}{\,h\,}=\dfrac 12.$
A: $$\frac{1}{h}\,\log\left(\frac{2+h}{2}\right) = \int_{0}^{1}\frac{dx}{2+hx}$$
hence by the dominated convergence theorem:
$$ \lim_{h\to 0} \frac{1}{h}\,\log\left(\frac{2+h}{2}\right) = \int_{0}^{1}\frac{dx}{2}=\color{red}{\frac{1}{2}}.$$
A: Set $h=2n$ to find $$\dfrac12\cdot\lim_{n\to0}\dfrac{\ln(1+n)}n=?$$
See Determine the following limit as x approaches 0: $\frac{\ln(1+x)}x$
A: For the first problem, we have
$$\lim\limits_{h\to 0}\frac{1}{h}\ln\left(\frac{2+h}{2}\right)$$
$$=\lim\limits_{h\to 0}\frac{\ln\left(2+h\right)-\ln(2)}{h}$$
$$=\lim\limits_{h\to 0}\frac{\frac{\mathrm d}{\mathrm dh}\ln\left(2+h\right)-\frac{\mathrm d}{\mathrm dh}\ln(2)}{\frac{\mathrm d}{\mathrm dh}h}$$
$$=\lim\limits_{h\to 0}\frac{\frac{1}{2+h}-0}{1}=\lim\limits_{h\to 0}\frac{1}{2+h}=\frac12$$
For the second problem, for every $k\in\mathbb N$, we have
$$\lim\limits_{x\to\infty}\frac{x^k}{e^x}=\lim\limits_{x\to\infty}\frac{\frac{\mathrm d^k}{\mathrm dx^k}x^k}{\frac{\mathrm d^k}{\mathrm dx^k}e^x}=\lim\limits_{x\to\infty}\frac{k!}{e^x}=0$$
Note that I simply applied L'Hôpital's rule to the original limit $k$ number of times.
