$\lim a^{b_n}=a^b$ if $\lim b_n = b$ Let $a \in \mathbb{R}$ and $\lim b_n = b$ (for a real sequence $b_n$). I am trying to show that $\lim a^{b_n}=a^b$ by using the definition of the convergence of a sequence, but I cannot see how to estimate $|a^{b_n} - a^b|$ against $|b_n - b|$ properly. 
 A: Hint: $a>0$, $a^{b_n}=\exp \left(b_n\log \left(a\right)\right)$. The function $f\left(x\right)=\exp \left(x\log\left(a\right)\right)$ is continue.
A: Let $\epsilon > 0$.  Let $f(x) = a^x$.  I'm assuming $a$ is positive.  We want to find an $N$ such that if $n \geq N$, $|a^{b_n}-a^b| < \epsilon$.
Note that $V = (\log_a(a^b-\epsilon),\log_a(a^b+\epsilon))$ is an open interval containing $b$ (because the logarithm is order preserving, this is the same as saying that $f(b) \in f(V) = (a^b - \epsilon, a^b + \epsilon)$.
We can find an $N$ such that $n \geq N$ implies $b_n \in V$.  Then $a^{b_n}$ is within $\epsilon$ of $a^b$, as required.
A: Here is a possibility assuming $a>1$, I'll leave other values up to you.
Given $\varepsilon>0$, write
$$\varepsilon'=\frac{\log\bigl(1+\frac\varepsilon{a^b}\bigr)}{\log a}\ .$$
Then $\varepsilon'>0$, so for sufficiently large $n$ we have
$$|b_n-b|<\varepsilon'\ .$$
This gives
$$a^{b_n}-a^b=a^b(a^{b_n-b}-1)<a^b(a^{\varepsilon'}-1)=\varepsilon$$
and
$$a^{b_n}-a^b>a^b(a^{-\varepsilon'}-1)
  =a^b\left(\frac1{1+\frac\varepsilon{a^b}}-1\right)
  =-\frac\varepsilon{1+\frac\varepsilon{a^b}}
  >-\varepsilon\ .$$
So
$$|a^{b_n}-a^b|<\varepsilon\ .$$
A: Herein, we present a way forward that relies only on an elementary pair of inequalities and the squeeze theorem.  We begin with a primer.

PRIMER:
In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the exponential function satisfies the inequalities
$$1+x\le e^x\le \frac{1}{1-x} \tag 1$$
for $x<1$


First we note that we can write
$$\begin{align}
a^{b_n}-a^b&=a^b\left(a^{b_n-b}-1\right)\\\\
&=a^b\left(e^{(b_n-b)\log(a)}-1\right)\\\\
\end{align}$$
We take $n$ large enough to guarantee that $(b_n-b)\log(a)<1$.  Then, we have
$$a^b(b_n-b)\log(a) \le a^b\left(e^{(b_n-b)\log(a)}-1\right)\le \frac{a^b(b_n-b)\log(a) }{1-(b_n-b)\log(a) } \tag 2$$
whereupon applying the squeeze theorem to $(2)$ we obtain the coveted limit

$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}(a^{b_n}-a^{b})=0}$$

And we are done!
A: $\{b_n\}$ converges to $b$
$n>N \implies |b_n - b| < \epsilon_1$
$a^{b_n} - a^b = a^b(a^{b_n-b}-1)$
$\{a^{b_n}\}$ converges to $a^b$ if there exists an $\epsilon_1$ such that $|(a^b)(1-a^{\epsilon_1})|<\epsilon$
$\epsilon_1 < \log_a (1-\epsilon a^{-b})$  
