squeeze theorem limit

Dec 2015
1,076
166
Earth
How can I prove that \(\displaystyle l=\lim_{x\rightarrow \infty} \dfrac{x^r } {e^x } =0 \; , r\in\mathbb{R}. \) using squeeze theorem (comparison ).
 
Last edited:
Dec 2015
1,076
166
Earth
\(\displaystyle \dfrac{x^r }{e^x } =\left(\frac{x}{e^{x/r}}\right)^{\!r}= r^r\left(\frac{x/r}{e^{x/r}}\right)^{\!r}<(2r)^r \sqrt{e^{-x}}. \)

\(\displaystyle 0< r^r\left(\frac{x/r}{e^{x/r}}\right)^{\!r}<(2r)^r \sqrt{e^{-x/r}}\) ; \(\displaystyle \; \; 0 < l <\lim_{x\rightarrow \infty }(2r)^r \sqrt{e^{-x/r}}=0 \implies l=0.\)

Also to confirm the result with l'hopital's rule : \(\displaystyle 0<x^r e^{-x} < x^{\lceil r \rceil } e^{-x}\) ; \(\displaystyle \; \; 0<l<\lim_{x\rightarrow \infty } \dfrac{(x^{\lceil r \rceil })^{(\lceil r \rceil)}}{(e^x)^{(\lceil r \rceil )}}=\lim_{x\rightarrow \infty } \dfrac{\lceil r \rceil !}{e^x } =0.\)
 
Last edited:

v8archie

Math Team
Dec 2013
7,710
2,679
Colombia
For $t > 1, \, c > 0$
\begin{alignat}{2}
0 &< t^{-1} &&< t^{c-1} \\
0 &< \int_1^x \frac1t \, \mathrm dt &&< \int_1^x t^{c-1}\, \mathrm dt \quad (x > 1) \\
0 &< \ln x &&< \tfrac1c (x^c - 1) < \tfrac1c x^c \\
0 &< \frac{\ln^a x}{x^b} &&< \tfrac1c x^{ac-b} \quad (a > 0, \, b > 0)
\end{alignat}
Now take $c = \frac{b}{2a} > 0$ so that we have
\[ 0 < \frac{\ln^a x}{x^b} < \tfrac1c x^{-\frac{b}{2}} \]

Finally, take limits as $x \to \infty$.

For your result, set $x = e^y$.
 
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