# Proving $\lim_{x\to \infty} xe^{-x^2} = 0$ using definition

I am trying to prove that $$\displaystyle\lim_{x\rightarrow \infty} xe^{-x^2} = 0$$ using the formal definition of limits as $$x$$ approaches infinity. The definition I am using is as follows:

$$\lim_{x\rightarrow \infty} f(x) = L$$ if the following is satisfied: for every number $$\varepsilon >0$$ there exists a number $$R$$, possibly depending on $$\varepsilon$$, such that if $$x>R$$, then $$x$$ belongs to the domain of $$f$$ and $$\vert f(x) - L \vert < \varepsilon.$$

My attempt has been to set up $$\vert xe^{-x^2} \vert < \varepsilon$$ and try to find a value of $$x$$ that satisfies the definition. However I am stuck. I tried rewriting it as $$e^{-x^2} < \frac{\varepsilon}{x}$$ so that $$x > \sqrt{\ln(\frac{\varepsilon}{x})}.$$ I am not able to solve for an explicit x-value though. I also tried writing $$x^2-\ln(x) > \varepsilon.$$ Is it possible to argue that any "nicer" expression than $$x^2-\ln(x)$$ could replace $$x^2-\ln(x)$$ and then find a value which $$x$$ must satisfy such that when $$x>R$$ we are guaranteed that $$\vert f(x) \vert< \varepsilon$$ ? Any input to this question would be appreciated!

• If you want to prove that $x^2-\ln x>\epsilon$ for big enough $x$, you can use the inequality $\ln x\geq x-1<x$(which comes from convexity) so, your problem is reduced to find big value $x$ with $x^2-x>\epsilon$, which should be easier since it's a polynomial. – Julian Mejia May 26 at 2:46

It seems you are specifically interested in getting an "easy" $$x_{\epsilon}$$ for given $$\epsilon > 0$$.

In such a case it is often helpful to estimate the given expression by others which are easier to handle. A possible way which plays it down to only elementary facts uses

• $$e > 2 \Rightarrow e^{-x} < 2^{-x}$$
• For $$x>1$$ we have $$e^{-x^2} < e^{-x}$$
• $$2^x \geq 2^{\lfloor x\rfloor} = (1+1)^{\lfloor x\rfloor} \stackrel{x>2}{>} \binom{\lfloor x\rfloor}{2}= \frac{\lfloor x\rfloor\cdot (\lfloor x\rfloor-1)}{2}$$

Let $$x>2$$ and $$\epsilon >0$$:

$$\frac{x}{e^{x^2}} < \frac{x}{2^x} < \frac{2 \lfloor x\rfloor}{\lfloor x\rfloor\cdot (\lfloor x\rfloor-1)} = \frac{2 }{ \lfloor x\rfloor-1} < \frac{2 }{ \lfloor x\rfloor}\stackrel{!}{<}\epsilon$$

Obviously, the requested inequality is satisfied for $$\boxed{\lfloor x_{\epsilon}\rfloor > \frac{2}{\epsilon}}$$

By $$e^{x^2}>1+x^2$$ then $$xe^{-x^2}<\dfrac{x}{1+x^2}<\dfrac1x$$

• So since $xe^{-x^2} >0$ for all $x>0$ it would suffice that I apply the epsilon-delta definition to $\frac{1}{x}$ and show that the limit is $0$ as $x$ approaches infinity? – Thomas Fjærvik May 26 at 1:32

Note that

$$xe^{-x^2} < xe^{-x}$$

for $$x>1$$. Working out the epsilon-delta argument for this function should be much easier.

For example $$e^x>\frac{x^2}{2}$$ since this is only one of the positive terms of the power series of $$e^x$$. It follows that $$\frac{x}{e^x}<\frac{2}{x}$$ so if you give me a positive $$\epsilon$$, $$xe^{-x}<\epsilon$$ for all $$x>\frac{2}{\epsilon}$$ QED.

• Then work it out... – Shalop May 26 at 2:20
• @Shalop There you have it! I said it was going to be easy! – plus1 May 26 at 2:52