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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!

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    $\begingroup$ 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. $\endgroup$ – Julian Mejia May 26 at 2:46
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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 $\boxed{\lfloor x_{\epsilon}\rfloor > \frac{2}{\epsilon}}$

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

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  • $\begingroup$ 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? $\endgroup$ – Thomas Fjærvik May 26 at 1:32
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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.

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

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