Show that $\lim\limits_{x \to 2} \bigg( 2^{x} \bigg)=4$ using epsilon-delta

Question

I have the following problem,

Show that $\lim\limits_{x \to 2} \bigg( 2^{x} \bigg)=4$ by finding a $\delta$ that guarantees that $2^{x}$ is within $\varepsilon$ of $4$ when $\varepsilon=0.01$

I want to prove that by using the epsilon delta proof, so I start as follows: \begin{align*} \left| f(x) - L \right| &< \varepsilon\\ \left| 2^{x} -4 \right| &< \varepsilon\ \end{align*}

but then I find myself completely confused... and I am also confused about the bit "guarantees that $2^{x}$ is within $\varepsilon$ of $4$ when $\varepsilon=0.01$", I am not sure what to do with $\varepsilon=0.01$.. I would really appreciate some help.. been stuck on this one for a while...

@DonThousand I was able to solve similar problems but only been stuck on this one, maybe I did not understand it as much as I thought I did — Friedrich, Feb 22 '20 at 15:22
Because the "guarantees that $2^x$ is within $\varepsilon$ of $4$" is the standard process in epsilon-delta proofs. I could explain how to do this one, but I'm of the opinion that this should tell you that you need to study the methods a bit better. — Rushabh Mehta, Feb 22 '20 at 15:23
@DonThousand I think my main problem is that I am not sure how to go from the first step, I think if I knew I would have been able to proceed — Friedrich, Feb 22 '20 at 15:25
@DonThousand I meant going from $\left| 2^{x} -4 \right| < \varepsilon $ — Friedrich, Feb 22 '20 at 15:31
Ok, here's a hint: Consider $\lim\limits_{x\to2^+}|2^x-4|$ first. So, $2^x-4>0$. So you can drop the absolute value. — Rushabh Mehta, Feb 22 '20 at 15:35
Then, when doing $\lim\limits_{x\to2^-}$, you simply consider $4-2^x$. — Rushabh Mehta, Feb 22 '20 at 15:36

score 1 · Accepted Answer · answered Feb 22 '20 at 16:08

You are trying to prove the following. For any given $\epsilon>0$, there exists $\delta>0$ such that when $0<|x-2|<\delta,$ it follows that $|2^x-4|<\epsilon$.

Consider when $x>2$. Since $2^x$ is increasing, $|2^x-4|=2^x-4$ so solving $2^x-4<\epsilon$, this holds iff $2^x<\epsilon+4$ iff $x<\log_2(\epsilon+4)$ iff $x-2<\log_2(\epsilon+4)-2$. The right hand side is positive since $\log_2(\epsilon+4)>\log_2(4)=2$. The if and only ifs allow you to go backwards in the proof. Therefore, for the right limit, it seems appropriate to choose $\delta_1=\log_2(\epsilon+4)-2$.

Now consider when $x<2$. Then it becomes $4-2^x<\epsilon$ iff $4-\epsilon<2^x$ iff $\log_2(4-\epsilon)<x$ iff $-\log_2(4-\epsilon)>-x$ iff $2-\log_2(4-\epsilon)>2-x$ . Thus, for the leftward limit, it seems that $\delta_2=2-\log_2(4-\epsilon)$ works.

For the combined limit, we want $0<|x-2|<\delta$. Choose $\delta=\min\{\delta_1,\delta_2\}$.

Show that $\lim\limits_{x \to 2} \bigg( 2^{x} \bigg)=4$ using epsilon-delta

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