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

**Pinkk** · March 16th 2010, 06:02 PM

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 $\lim_{n\to \infty} x_{n} = x$ , then $\lim_{n\to \infty} 2^{x_{n}} = 2^{x}$ . 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.

**Drexel28** · March 16th 2010, 06:05 PM

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 $\lim_{n\to \infty} x_{n} = x$ , then $\lim_{n\to \infty} 2^{x_{n}} = 2^{x}$ . 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 $2^x-2^y$ ? First prove that $f$ is continuous at zero and then note that $|2^x-2^y|=|2^y||2^{\frac{x}{y}}-1|$

**Pinkk** · March 16th 2010, 06:14 PM

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

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

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

I think that's correct, right?

**Drexel28** · March 16th 2010, 06:32 PM

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 $f(x)=2^{x}$ is continuous on $(0,1)$ ? Because then I want to show that there exists delta such that $|x-x_{0}|<\delta \implies |2^{x} - 2^{x_{0}}|< \epsilon$ for $\epsilon > 0, x\in (0,1)$

Forget $(0,1)$ .

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

Proof:

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

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

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

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

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

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

**Drexel28** · March 16th 2010, 10:37 PM

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 $\varphi:\mathbb{R}\to\mathbb{R}$ such that $\varphi(x)\varphi(y)=\varphi(x+y)$ (and thus the only function of this form which is continuous at $0$ as I prove above) is $\varphi(x)=\varphi(1)^x$ .

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

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

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