Some advanced properties of limit (on sequences) I have spent too much time looking for answers of these problem by myself. I am wondering about the ways to solve:
Suppose sequence $\{a_n\}$ converges to $k$ and $\{b_n\}$ converges to $l$.


*

*if $c$ is a number, prove that $\lim_{n\to\infty} c^{a_n}=c^{\lim_{n\to\infty} a_n}=c^k$.

*if $c>0$ and $l>0$, prove that $\{\log_c{b_n}\}$ converges to $\log_c{l}$.

*if $k\neq 0$ or $l\neq 0$ prove that $\{a_n^{b_n}\}$ converges to $k^l$.


Thanks a lot if you help me :")
 A: I will use without proof in the following, that $\exp:\mathbb{R}\to\mathbb{R}$ and $\ln:(0,\infty)\to\mathbb{R}$ are continuous.
Let $(a_n), (b_n)$ be sequences in $(0,\infty),\mathbb{R}$ respectively with $\lim_{n\to\infty}a_n=a,\lim_{n\to\infty}b_n=b$. Suppose $c,a\in (0,\infty)$. Note that these prerequisites differ slightly from your more unspecific statements, as we have to be cautious with the logarithm and negative numbers.

As $\ln$ is continuous(as $\exp$ is continuous), we derive the following over a change of basis in the logarithm:
$$
\lim_{n\to\infty}\log_ca_n=\lim_{n\to\infty}\frac{\ln a_n}{\ln c}=\frac{1}{\ln c}\lim_{n\to\infty}\ln a_n=\frac{1}{\ln c}\ln(\lim_{n\to\infty}a_n)=\frac{\ln a}{\ln c}=\log_ca
$$

As $\exp$ and $\ln$ are continuous, we derive that
$$
\lim_{n\to\infty}a_n^{b_n}=\lim_{n\to\infty}\exp(\ln(a_n)\cdot b_n)=\exp(\lim_{n\to\infty}\ln(a_n)\cdot\lim_{n\to\infty}b_n)=\exp(\ln(a)\cdot b)=a^b
$$
The modified result for $c\in (0,\infty)$ follows essentially from this, taking $a_n:= c$.
