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$\lim_{x \to \infty} \frac{x^a}{e^x} = 0$

How do I prove this? I didn't get anywhere so I don't even think it is worth writing down how far I've. I've been stuck for a long time and would like help. And BTW no I've not learned l'hopital's rule yet (Someone suggested I solve it using it). Is there another way?

Dave L. Renfro
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Need_MathHelp
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    What tools do you have available? You can easily say that $e^x>1+x+x^2/2+...+x^{a+1}/(a+1)!$ if thats how you define $e^x$ – QC_QAOA Feb 03 '23 at 16:48
  • @QC_QAOA But then what? – Andrew Feb 03 '23 at 16:57
  • Step 1: Write the limit as $$\lim_{x\to\infty}\left(\frac{x}{e^{x/a}}\right)^a$$. So it suffices to show $$\lim_{x\to\infty}\frac{x}{e^{x/a}}=0$$ Step 2: Compute the integral $$\int_0^{\infty}xe^{-x/a},dx$$ Step 3: Conclude – Andrew Feb 03 '23 at 17:02

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We can prove this step by step, beginning with a sequences. Consider the sequence $\frac{n^a}{b^n}$, where $b>1$ and $a>0$ are real numbers. Let $b=1+\lambda$, where $\lambda>0$. Then by binom formula we have $$b^n= (1+\lambda)^n=1+n\lambda+\frac{n(n-1)}{2}\lambda^2+...>$$ $$>\frac{n(n-1)}{2}\lambda^2. $$ Because for $n>2$ we have $n-1>\frac{n}{2}$, we have $$b^n>\frac{(b-1)^2}{4}n^2.$$ For $a=1$ we get that $\frac{n}{b^n}<\frac{4}{(b-1)^2}\cdot \frac{1}{n}$ and $$\lim\limits_{n\to\infty}\frac{n}{b^n}=0.$$ Since this result is correct for all $b>1$, then taking $a>1$ we have $$\frac{n^a}{b^n}=\left( \frac{n}{(b^{1/a})^n} \right)^a< \frac{n}{(b^{1/a})^n}.$$ Therefore, $$\lim\limits_{n\to\infty} \frac{n^a}{b^n}=0.$$ Proved thus for $a\ge 1$, this result is all the more true for $a<1$.

Now let's go to the limit of the function. From previous we already know that $\lim\limits_{n\to\infty}\frac{n}{b^n}=0$. Then $\lim\limits_{n\to\infty}\frac{n+1}{b^n}=0$.

By the definition for given $\epsilon>0$ there exists natural number $N$ such that for all $n>N$ we have $\frac{n+1}{b^n}<\epsilon$.

Now let $x>N+1$. If we put $n=[x]$, then $$ n>N \text{ and } n\le x<n+1,$$ so $$\frac{x}{b^x}<\frac{n+1}{b^n}<\epsilon$$ and $\lim\limits_{x\to+\infty}\frac{x}{b^x}=0$.

Similarly, from here we get that for $a>1$ $$\frac{x^a}{b^x}=\left( \frac{x}{(b^{1/a})^x} \right)^a< \frac{x}{(b^{1/a})^x}.$$ Hence, $$\lim\limits_{x\to+\infty}\frac{x^a}{b^x}=0.$$ Obviously it's true also for $a<1$.

For $b=e$ we get our result.

greyls
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