Limit of a sequence of real numbers If $(a_n), (b_n)$ are two sequences of real numbers so that $(a_n)\rightarrow a,\,\,(b_n)\rightarrow b$ with $a, b\in \mathbb{R}^+$.  How to prove that $a_n^{b_n}\rightarrow a^b$ ?
 A: Since $a_n\to a$ and $a>0$ by assumption, we have $a_n>0$ for $n\geq N$ for some sufficiently large positive integer $N$. So we can just consider $\log a_n$ for $n\geq N$. Note that $\log$ is a continuous function, we have 
$$\lim_{n\to\infty}(\log a_n)=\log(\lim_{n\to\infty} a_n)=\log a.$$
Therefore, we have
$$\log\Big(\lim_{n\to\infty} a_n^{b_n}\Big)=\lim_{n\to\infty}(\log a_n^{b_n})=\lim_{n\to\infty}(b_n\log a_n)=(\lim_{n\to\infty}b_n)(\lim_{n\to\infty}\log a_n)=b\log a=\log a^b,$$
which implies that (by taking exponential on both sides)
$$\lim_{n\to\infty} a_n^{b_n}=a^b$$
as required.
A: Note: The statement doesn't require $b > 0$.  We don't assume it here.
If we take the continuity of $\ln$ and $\exp$ for granted, the problem essentially boils down to showing $b_n x_n \to bx$ where $x_n = \ln a_n, x = \ln a$ since $a_n^{b_n} = e^{b_n x_n}$.  This is what the work in the first proof below goes toward (the "add-and-subtract $b_n \ln a_n$" trick comes up elsewhere too).  But if you can use $b_n \to b \text{ and } x_n \to x \implies b_n x_n \to bx$ then the proof simplifies to the second one below.

Proof:  Given $\varepsilon > 0$, there exist $K_1, K_2 \in \mathbb{N}$ such that for $n \in \mathbb{N}$ we have $n > K_1 \implies |\ln a_n - \ln a| < \frac{\varepsilon}{2(|b|+1)}$ (since $a_n \to a \in \mathbb{R}^+ \implies \ln a_n \to \ln a$ by continuity of $\ln$ on $\mathbb{R}^+$) and
  $n > K_2 \implies |b_n - b| < \min(\frac{\varepsilon}{2 (|\ln a| + 1)},1)$ (by hypothesis).  Let $K = \max(K_1, K_2)$.  Then
  $$\begin{eqnarray}
n > K \implies |b_n \ln a_n - b \ln a| &=& |b_n \ln a_n - b_n \ln a + b_n \ln a - b \ln a|\\
&\leq& |b_n \ln a_n - b_n \ln a| + |b_n \ln a - b \ln a|\\
&=& |b_n|\,|\ln a_n - \ln a| + |b_n - b|\,|\ln a|\\
&<& (|b| + 1)\,\frac{\varepsilon}{2(|b|+1)} + \frac{\varepsilon}{2 (|\ln a|+1)}\,|\ln a|\\
&<& \varepsilon
\end{eqnarray}$$
  so $b_n \ln a_n \to b \ln a \in \mathbb{R}$ by definition.  But $x \mapsto e^x$ is continuous on $\mathbb{R}$, hence $a_n^{b_n} = e^{b_n \ln a_n} \to e^{b \ln a} = a^b$.

As mentioned at the top, using the theorem on the limit of a product gives this:

Short proof:
  $$\begin{eqnarray}
a^b &=& \exp\left[b \ln a\right]\\
&=& \exp\left[\left(\lim_{n \to \infty} b_n\right)\left(\ln \lim_{n \to \infty} a_n\right)\right] &\text{ assumption}&\\
&=& \exp\left[\left(\lim_{n \to \infty} b_n\right)\left(\lim_{n \to \infty} \ln a_n\right)\right] &\text{ continuity of }\ln\text{ at }\lim_{n \to \infty}a_n = a \in \mathbb{R}^+&\\
&=& \exp\left[\lim_{n \to \infty} b_n \ln a_n\right] &\text{ product of limits }\longrightarrow\text{ limit of product}&\\
&=& \lim_{n \to \infty} e^{b_n \ln a_n} &\text{ continuity of }\exp\text{ at }\lim_{n \to \infty}b_n \ln a_n = b \ln a \in \mathbb{R}&\\
&=& \lim_{n \to \infty} a_n^{b_n}
\end{eqnarray}$$

All the work and $\varepsilon$'s are still there, but now they're hidden in the theorems we used.
A: The function $f(x,y) = x^y = e^{y \ln x}$ is continuous on $\mathbb{R}_+ \times \mathbb{R}$, hence if $(a_n,b_n) \to (a,b)$ (with $a_n, a >0$, of course), then $a_n^{b_n} = f(a_n,b_n) \to f(a,b) = a^b$.
