Prove that if a sequence $(a_n)$ converges to some limit $L$, then $2^{a_n}$ converges to $2^L$ I'm completely lost here. 
Attempt
We know that $a_n$ converges to $L$. So by the definition convergence, $\forall \varepsilon \in R, \exists M \in N$ such that $|a_n - L| < \varepsilon, \forall n \ge M$.
We need to use this fact to show that $\forall \varepsilon \in R, \exists M' \in N$ such that $|2^{a_n} - 2^L| < \varepsilon, \forall n \ge M'$
I'm completely lost as to where to start. I can't think of any properties of exponents that I can exploit here. Would appreciate some hints.
Thank You
 A: Hint: $2^x$ is a continuous function (can you show this?). For a continuous function $f$ we have $f(x_n) \to f(x)$ as $x_n \to x$.
Another way: If you wanted to do $\varepsilon-\delta$ language then you could also consider $|2^{a_n}-2^L|=2^L|2^{a_n-L}-1|$
A: We want \begin{align}
|2^{a_n}-2^L| &= 2^L |2^{a_n - L} - 1|< \epsilon
\end{align}
$$|2^{a_n-L}-1|  < \frac{\epsilon}{2^L}$$
$$1-\frac{\epsilon}{2^L}<2^{a_n -L}< 1+\frac{\epsilon}{2^L}$$
Focus on small $\epsilon$ to make sure the lower bound is positive
$$\log _2\left(1-\frac{\epsilon}{2^L}\right)<{a_n -L}< \log_2\left( 1+\frac{\epsilon}{2^L}\right)$$
Hence we want to choose $n$ large enough that 
$$|a_n-L| < \min\left(-\log _2\left(1-\frac{\epsilon}{2^L}\right),  \log_2\left( 1+\frac{\epsilon}{2^L} \right)\right) $$
A: A minor point of clarification in your post: just to point out that it is not the case that for any ∀ε∈R,∃M∈N such that the conditions hold. This is because ε must be positive.
A: $$\lim 2^{a_n} = 2^{\lim a_n} = 2^L$$
I suggest to not use $\varepsilon-N$/$\varepsilon-\delta$ here because you would be reinventing the wheel: Instead use $\varepsilon-N$/$\varepsilon-\delta$ to prove that for a continuous function $f$, $$\lim f(a_n) = f(\lim a_n).$$
Otherwise, you'd use $\varepsilon-N$/$\varepsilon-\delta$ every time?
