# Thread: Prove f(x)=x2^x is continuous on (0,1)

1. ## Prove f(x)=x2^x is continuous on (0,1)

I'm blanking out, how would I show either a sequential limit or epsilon-delta proof for this? For the sequential limit definition, is it an obvious property that if $\displaystyle \lim_{n\to \infty} x_{n} = x$, then $\displaystyle \lim_{n\to \infty} 2^{x_{n}} = 2^{x}$. If not, how do I go about proving that, or how would I show an epsilon-delta proof? I am completely stuck on how it could be shown using epsilon-delta. Thanks.

2. Originally Posted by Pinkk
I'm blanking out, how would I show either a sequential limit or epsilon-delta proof for this? For the sequential limit definition, is it an obvious property that if $\displaystyle \lim_{n\to \infty} x_{n} = x$, then $\displaystyle \lim_{n\to \infty} 2^{x_{n}} = 2^{x}$. If not, how do I go about proving that, or how would I show an epsilon-delta proof? I am completely stuck on how it could be shown using epsilon-delta. Thanks.
....isn't the multiplication of two continuous functions continuous. Or do you mean $\displaystyle 2^x-2^y$? First prove that $\displaystyle f$ is continuous at zero and then note that $\displaystyle |2^x-2^y|=|2^y||2^{\frac{x}{y}}-1|$

3. I'm sorry, I don't see how that helps. Are you trying to use that for an epsilon delta proof that $\displaystyle f(x)=2^{x}$ is continuous on $\displaystyle (0,1)$? Because then I want to show that there exists delta such that $\displaystyle |x-x_{0}|<\delta \implies |2^{x} - 2^{x_{0}}|< \epsilon$ for $\displaystyle \epsilon > 0, x\in (0,1)$

Err, nevermind, I caught your error and with the correction I understand it.

Okay, so I worked out if $\displaystyle \delta = log_{2}(\frac{\epsilon}{2^{x_{0}}} +1)$, then with manipulation we get if $\displaystyle |x-x_{0}| <\delta$ then $\displaystyle |2^{x} - 2^{x_{0}}|< \epsilon$

I think that's correct, right?

4. Originally Posted by Pinkk
I'm sorry, I don't see how that helps. Are you trying to use that for an epsilon delta proof that $\displaystyle f(x)=2^{x}$ is continuous on $\displaystyle (0,1)$? Because then I want to show that there exists delta such that $\displaystyle |x-x_{0}|<\delta \implies |2^{x} - 2^{x_{0}}|< \epsilon$ for $\displaystyle \epsilon > 0, x\in (0,1)$
Forget $\displaystyle (0,1)$.

Theorem: Let $\displaystyle f:\mathbb{R}\to\mathbb{R}$ be such that $\displaystyle f$ is continuous at $\displaystyle 0$ and $\displaystyle f(x)f(y)=f(x+y)$ then $\displaystyle f$ is continuous on $\displaystyle \mathbb{R}$.

Proof:

If $\displaystyle f(0)=0$ we're done since $\displaystyle f(x)=f(x+0)=f(x)f(0)=f(x)0=0$ and $\displaystyle f(x)=0$ is trivially continuous. Thus, assume that $\displaystyle f(0)\ne 0$ then $\displaystyle f(0)=f(0+0)=f(0)f(0)\implies f(0)=1$. Also, notice that $\displaystyle f(x)\ne 0$ for any $\displaystyle x\in\mathbb{R}$ since we covered the case above assume that $\displaystyle x\ne 0$ then $\displaystyle f(x)=0\implies f(x)f(-x)=0\implies f(0)=0$ which is obviously a contradiction. It follows that $\displaystyle f(x)f(-x)=f(0)=1\implies f(-x)=\frac{1}{f(x)}$ and so $\displaystyle \frac{f(x)}{f(y)}=f(x-y)$

Thus, since $\displaystyle f$ is continuous at $\displaystyle 0$ for every $\displaystyle \varepsilon>0$ there exists some $\displaystyle \delta>0$ such that $\displaystyle |x|<\delta\implies |f(x)-1|<\varepsilon$.

So, let $\displaystyle \varepsilon>0$ be given and $\displaystyle x_0\in\mathbb{R}$ arbitrary. Then, $\displaystyle |f(x_0)-f(x)|=|f(x_0)|\left|1-\frac{f(x)}{f(x_0)}\right|=|f(x_0)|\left|1-f(x-x_0)\right|$

Now, let $\displaystyle z=x-x_0$. We know that there exists a $\displaystyle \delta>0$ such that $\displaystyle |z|<\delta\implies |1-f(z)|<\frac{\varepsilon}{|f(x_0)|}$. Thus, if $\displaystyle |x-x_0|=|z|<\delta\implies |f(x_0)||1-f(x-x_0)|=|f(x_0)||1-f(z)|<|f(x_0)|\frac{\varepsilon}{|f(x_0)|}=\vareps ilon$. It follows that $\displaystyle f(x)$ is continuous at $\displaystyle x_0$.

Since $\displaystyle x_0$ was arbitrary it follows that $\displaystyle f(x)$ is continuous on $\displaystyle \mathbb{R}$. $\displaystyle \blacksquare$.

All you have to do is prove that $\displaystyle f(x)=2^x$ is continuous at $\displaystyle 0$

5. It might be interesting to point out that while I showed the proof I did because it is interesting it turns out that the only continuous function $\displaystyle \varphi:\mathbb{R}\to\mathbb{R}$ such that $\displaystyle \varphi(x)\varphi(y)=\varphi(x+y)$ (and thus the only function of this form which is continuous at $\displaystyle 0$ as I prove above) is $\displaystyle \varphi(x)=\varphi(1)^x$.

To see this first note that by induction we have that for every $\displaystyle n\in\mathbb{N}$ that $\displaystyle \varphi(n)=\varphi(\underbrace{1+\cdots+1}_{n\text { times}})=\underbrace{\varphi(1)\cdots\varphi(1)}_{ n\text{ times}}=\varphi(1)^n$. Therefore, if $\displaystyle n\in\mathbb{N}$ we have that $\displaystyle \varphi(1)=\varphi\left(\frac{n}{n}\right)=\varphi \left(\frac{1}{n}\right)^n\implies \varphi^{\frac{1}{n}}=\varphi\left(\frac{1}{n}\rig ht)$. THus, we have that if $\displaystyle \frac{p}{q}\in\mathbb{Q}$ that $\displaystyle \varphi\left(\frac{p}{q}\right)=\varphi\left(\frac {1}{q}\right)^p=\varphi(1)^{\frac{p}{q}}$.

It follows that if we define $\displaystyle A\left(\varphi(x),\varphi(1)^x\right)=\left\{y\in\ mathbb{R}:\varphi(y)=\varphi(1)^y\right\}$ we have that $\displaystyle A(\varphi(x),\varphi(1)^x)$ is dense in $\displaystyle \mathbb{R}$ and since it's closed it follows that $\displaystyle A(\varphi(x),\varphi(1)^x)=\overline{A(\varphi(x), \varphi(1)^x)}= \mathbb{R}$ from where the conclusion follows.