how to prove lim [x->infty][p(x) / e^x] = 0 for p(x) = ..

[ku1005]

Hi, with the following Q, i think I know why , but im not quite sure how t prove??Anyways, any suggestions would be great!

Let



Where a0, a1, ..., a(n-1) are constant real numbers. Prove above for n <= 3.

Give your reasoning. What happens for n > 3?

...............................................

My understanding is that e^x increases at a greater rate then any polynomial of degree less then or equal to 3. So thus, if n>3, the limit should tend towards infinity, however im not sure how to prove this.

 

[pka]

Question
Do you have l’Hopital’s rule to use?
If not what do you know about bounds?

 

[galactus]

You could rewrite your limit as:

\(\displaystyle \L\\\lim_{x\to\infty}\frac{x^{n}}{e^{x}}+\lim_{x\to\infty}\frac{a_{n-1}x^{n-1}}{e^{x}}+\lim_{x\to\infty}\frac{a_{n-2}x^{n-2}}{e^{x}}+......+\lim_{x\to\infty}\frac{a_{1}x}{e^{x}}+\lim_{x\to\infty}\frac{a_{0}}{e^{x}}\)

One would think, since e^x increases without bound, the limit tends to 0.

If you had \(\displaystyle \L\\\lim_{x\to\infty}\frac{x^{3}}{e^{x}}+\lim_{x\to\infty}\frac{a_{2}x^{2}}{e^{x}}+\lim_{x\to\infty}\frac{a_{1}x}{e^{x}}+\lim_{x\to\infty}\frac{a_{0}}{e^{x}}\)

You can see that since \(\displaystyle \L\\\lim_{x\to\infty}\frac{x^{3}}{e^{x}}=0\) the other terms do also.

Maybe this isn't viable enough of a proof, but you can see what I mean.

 

[pka]

Here is a formal proof.
\(\displaystyle \L e^x = \sum\limits_{k = 0}^\infty {\frac{{x^k }}{{k!}}} \quad \Rightarrow \quad \frac{{e^x }}{{x^n }} = \sum\limits_{k = 0}^\infty {\frac{{x^{k - n} }}{{k!}}} = \sum\limits_{k = 0}^n {\frac{{x^{k - n} }}{{k!}}} + \sum\limits_{k = 1}^\infty {\frac{{x^k }}{{\left( {n + k} \right)!}}}\)

From that bit of algebra we see that \(\displaystyle \L \lim _{x \to \infty } \frac{{e^x }}{{x^n }} = \infty \quad \left( {\forall n \in Z^ + } \right)\).

Then \(\displaystyle \L \left( {\forall n \in Z^ + } \right)\left[ {\lim _{x \to \infty } \frac{{x^n + a_{n - 1} x^{n - 1} + \cdots + a_1 x + a_0 }}{{e^x }}} \right] = \lim _{x \to \infty } \frac{{1 + a_{n - 1} x^{ - 1} + \cdots + a_1 x^{1 - n} + \frac{{a_0 }}{{x^n }}}}{{\frac{{e^x }}{{x^n }}}} = 0\).

 

[ku1005]

thanks for all your help guys, but after thinking about it for a little bit, I decided that L'Hopitals rule would be the most useful and thus used that in my proof...thnaks again