Prove that $x^m/e^x\to0$ when $x\to+\infty$, for every $m$

Question

Defn: We write $\lim_{x \uparrow \infty} f(x)=L$ whenever $f$ is defined on some unbounded interval such as $0<x<\infty$ and, corresponding to any $\epsilon >0$. $\exists x_0$ such that $\mid f(x) - L \mid < \epsilon$ whenever $x>x_0$

Prove that for any constants m, $\lim_{x \uparrow \infty} \frac{x^m}{e^x}=0$

Well my problem is I was working on its definition by working backwards, i.e starting at $\mid \frac{x^m}{e^x} - 0 \mid < \epsilon$ to find $x_0$. Tried working at cases by letting $m=0$ in which I found $x_0 = ln(\frac{1}{\epsilon})$. I did this to avoid unnecessary "undefineded" when I divide $m$.

Here's my case, I can't seem to find a proper $x_0$ when I pick $m\neq 0$. Well take a look

$\mid \frac{x^m}{e^x} - 0 \mid = \mid \frac{x^m}{e^x}\mid < \epsilon$

$\Rightarrow \mid x^m \mid < \epsilon e^x$

$\Rightarrow x^m < \epsilon e^x$ since x is from 0 to infinity, I can drop the absolute sign.

$\Rightarrow \frac{1}{\epsilon} < e^x \frac{1}{x^m}$

$\Rightarrow ln(\frac{1}{\epsilon}) < x - ln (x^m)$

in which I have no idea how to combine $x$ in a single term such that $x>x_0$ Please help.

Answer 1

What you might want to use is how $e^x$ will grow faster than $x^m$ regardless of $m$. Let's see what happens if we double $x$:

$$e^{2x} = (e^x)^2$$ $$(2x)^m = 2^m x^m$$

That is when doubling $x$ the denominator will grow with a factor $e^x$ while the nominator will grow with a factor $2^m$. This gives us a a point where the denominator will start to grow faster than the nominator (by a factor 2), that is when $e^x > 2^{m+1}$. This fact can then be used to estimate ratio to be $1/2$ of the original each time $x$ is doubled.

So let's put it together:

Let $a = \ln2^{m+1}$ and consider the intervals $I_j=[2^{j}a, 2^{j+1}a]$ (noting that these intervals will cover all $x>a$). We prove that for $x\in I_j$ we have $x^m \le (2a)^m 2^{jm}$ and $e^x \ge 2^{(m+1)j}$ giving $x^m/e^x \le (2a)^m/2^j$. This is done using induction:

For $j=0$ we have $x \le 2a$ which means that $x^m < (2a)^m = (2a)^m 2^{jm}$.

If we assume that the estimate is true for some $j-1 \ge 0$ we have for $x\in I_j$ that $x^m = (2 (x/2))^m = 2^m (x/2)^m$, but $x/2\in I_{j-1}$ so $x^m = 2^m (x/2)^m \le 2^m (2a)^m2^{(j-1)m} = (2a)^m2^{jm}$.

Similarly one can prove that for $x\in I_j$ that $e^x\ge2^{(m+1)j}$, which is left as an exercise.

The proof is then concluded by selecting $k$ such that $x^m/e^x < \epsilon$ for $x\in I_j$ for all $j\ge k$.

Answer 2

Suppose we know that (either via L'Hospital's rule or some other method)

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

It follows for $m>0$ that by letting $t=x/m$ we get

$$\frac1m\lim_{x\to\infty}\frac x{e^{x/m}}=0\implies\lim_{x\to\infty}\frac x{e^{x/m}}=0$$

By continuity, we may raise both sides to the power of $m$ to get

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

as expected.

Of course, for $m<0$, it is easy to show that $x^m\to0$ so the limit is trivially obvious.

Answer 3

Without loss of generality, let $m > 0$. Note that $x^{m}/e^{x} = (\log t)^{m}/t$ for suitable $x,t$; so it suffices deal with the right hand side. If $t \geq 1$, then for all $c > 0$ we have $$ \frac{x^{m}}{e^{x}} = \frac{(\log t)^{m}}{t} = \frac{(\int_{1 \leq u \leq t}\frac{1}{u})^{m}}{t} \leq \frac{(\int_{1\leq u \leq t}u^{c-1})^{m}}{t} = \frac{(t^{c}-1)^{m}}{ct} < \frac{t^{cm}}{ct}; $$ hence if $c := 1/2m$, then $$ \frac{t^{cm}}{ct} = \frac{2m}{\sqrt{t}}. $$ If $\varepsilon > 0$, then $t > 4m^{2}/\varepsilon^{2}$ only if $2m/\sqrt{t} < \varepsilon$; hence for every $\varepsilon > 0$, we have $t > \max \{ 1, 4m^{2}/\varepsilon^{2} \}$ only if $(\log t)^{m}/t < \varepsilon$.

In fact, we do not have to stop here; the statement you want to prove can generalize to this statement: for all $a,b > 0$ we have $x^{a}/e^{bx} \to 0$ as $x$ grows indefinitely; I believe you can now see a why by applying the argument above.

Answer 4

If $m\leq 0$ the assertion is obvious. Assume therefore that $m>0$. Since for every $x>0$

$$e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!} > \frac{x^{m+1}}{(m+1)!}$$

we have for every $m>0$ and $x\neq 0$:

$$\frac{e^x}{x^m}>\frac{x}{(m+1)!}$$

whence the limit of the l.h.s as $x\to\infty$ is $\infty$. It follows that the limit of the reciprocal, $\frac{x^m}{e^x}$, as $x\to\infty$, is zero.

Answer 5

Let $f=x^me^{-x}$ so $\ln f=m\ln x -x$ and $(\ln f)'=\frac{m}{x}-1$. For $x>2m$, $(\ln f)'<-\frac{1}{2}$ so $\ln f<m\ln 2m-2m-\frac{1}{2}(x-2m)=-\frac{x}{2}+m\ln 2m-m$. Since $\ln f$ becomes arbitrarily negative as $x\to\infty$, $f\to 0^+$.