# Prove convergence of limit in $\mathbb{R}$.

Suppose $$a_n \to L$$ in $$\mathbb{R}$$. Prove $$2^{a_n} \to 2^L$$.

I tried to start from $$| 2^{a_n} - 2^L | < \epsilon$$ and work my way to $$|a_n - L| < \epsilon$$, but logarithms are showing up and I think the problem is too simply to need to prove an inequality with a logarithm or exponential.

I cannot use the continuity of $$2^n$$ in this question.

Any help is appreciated.

• Why did you downvote? At least comment and give a reason... – libby Oct 22 '18 at 13:40
• Can we use the fact that $f(x)=e^x$ is continuous? – Yanko Oct 22 '18 at 13:40
• No, we can't use continuity, sorry I should have added that in. – libby Oct 22 '18 at 13:41
• – Chinnapparaj R Oct 22 '18 at 16:15

Let $$\varepsilon>0$$. Since $$a_n\rightarrow L$$ we can choose $$N$$ such that for all $$n>N$$ we have $$|a_n-L|<\log_2(1+\frac{\varepsilon}{2^L})$$

Now for this $$N$$ we have for all $$n>N$$

$$|2^{a_n}-2^L| = |2^{a_n - L + L}-2^L|=2^L |2^{a_n-L}-1|$$

Since $$|a_n-L|<\log_2(1+\frac{\varepsilon}{2^L})$$ we have that $$|2^{a_n-L}|<1+\frac{\varepsilon}{2^L}$$ which means that $$1-\frac{\varepsilon}{2^L}<2^{a_n-L}<1+\frac{\varepsilon}{2^L}$$ and so by the above equation $$-\varepsilon <2^{a_n}-2^L <\varepsilon$$, equivalently $$|2^{a_n}-2^L |<\varepsilon$$. This completes the proof.

• Why can we take the absolute value in the last step? – libby Oct 22 '18 at 13:56
• @Libby Hmm you are right I only written one of the sides, but the other side also holds of course. Let me fix that – Yanko Oct 22 '18 at 15:13

Let $$e^b = 2$$, $$b>0$$.

$$|e^{ba_n} -e^{bL}|=$$

$$e^{bL}|e^{b(a_n-L)}-1|<$$

$$e^{bL}(2|b(a_n-L)|)$$ for

$$|b(a_n-L)| <1$$.

Used : $$|e^x-1|<2|x$$| for $$|x| <1$$.

1) Let $$\epsilon >0$$, for $$\dfrac{\epsilon}{e^{bL}2b+\epsilon} >0$$

there is a $$n_0$$ s.t. for $$n \ge n_0$$ :

$$|a_n-L| < \dfrac{\epsilon}{e^{bL}2b+\epsilon}<1$$,

2) $$|e^{ba_n}-e^{bL}|

$$(e^{bL}2b)\dfrac{\epsilon}{e^{bL}2b +\epsilon} <\epsilon.$$

Dividing by $$2^L$$ reduces this to the case when $$L = 0$$. So we want to show that $$a_n \to 0 \implies 2^{a_n} \to 1$$.

As usual, Bernoulli to the rescue!

$$(1+1/n)^n \ge 1+\frac{n}{n} = 2$$ so $$2^{1/n} \le 1+\frac1{n}$$.

Since $$2^{1/n} > 1$$, if $$2^x$$ is monotonic increasing, this gives us what we want. That, in turn, follows from $$2^x > 0$$ for all real $$x$$ and $$2^{x+h} = 2^x 2^h > 2^x$$.

Note that, more generally, if $$c > 0$$ then

$$(1+c/n)^n \ge 1+\frac{nc}{n} =1+c$$ so $$(1+c)^{1/n} \le 1+\frac{c}{n}$$.