# Math Help - Limits

1. ## Limits

Calculate the following limits:

1.) $\lim_{x \to \infty} \frac{\ln{x}}{x^{\frac{2}{3}}}$

I know the answer is 0. Can I simply use L'Hopitals?

So, $\frac{\frac{1}{x}}{\frac{2}{3}x^{\frac{-1}{3}}}$? Something like that?

And then,

2.) Does $\int_{1}^{\infty} \frac{e^{-x}}{\sqrt{x}}\, dx$ converge? Justify your answer.

No idea how to justify, but my calculator gives .2788 as an answer so it def does converge.

2. Originally Posted by Ideasman
Calculate the following limits:

1.) $\lim_{x \to \infty} \frac{\ln{x}}{x^{\frac{2}{3}}}$

I know the answer is 0. Can I simply use L'Hopitals?

So, $\frac{\frac{1}{x}}{\frac{2}{3}x^{\frac{-1}{3}}}$? Something like that?

And then,

2.) Does $\int_{1}^{\infty} \frac{e^{-x}}{\sqrt{x}}\, dx$ converge? Justify your answer.

No idea how to justify, but my calculator gives .2788 as an answer so it def does converge.
Yes you can use l'hopitals

3. Hello,

For the first one, i guess it's correct.

For the second one : if the function in the integral is continuous and the limits at the extremities of the integral are finite, then the integral converges.

In $[1;+\infty[$, $\frac{e^{-x}}{\sqrt{x}}$ is continuous.
In 1, the function has a finite value.

Now let's study the limit at $+\infty$

$\frac{e^{-x}}{\sqrt{x}}=\frac{1}{e^x \sqrt{x}}$

The limit is clearly 0.

So the integral converges.

I hope i didn't miss any point :s

4. Originally Posted by Ideasman
Calculate the following limits:

1.) $\lim_{x \to \infty} \frac{\ln{x}}{x^{\frac{2}{3}}}$

I know the answer is 0. Can I simply use L'Hopitals?

So, $\frac{\frac{1}{x}}{\frac{2}{3}x^{\frac{-1}{3}}}$? Something like that?

And then,

2.) Does $\int_{1}^{\infty} \frac{e^{-x}}{\sqrt{x}}\, dx$ converge? Justify your answer.

No idea how to justify, but my calculator gives .2788 as an answer so it def does converge.
since imputting ∞ into the top and bottom gives $\frac{\infty}{\infty}$...you can use L'hopital's rule...so $\lim_{x \to {\infty}}\frac{\ln(x)}{x^{\frac{2}{3}}}=\lim_{x \to {\infty}}\frac{\frac{1}{x}}{\frac{2x^{\frac{1}{3}} }{3}}=\lim_{x \to {\infty}}\frac{2}{3x^{\frac{4}{3}}}=0$

5. Originally Posted by Ideasman
Calculate the following limits:
1.) $\lim_{x \to \infty} \frac{\ln{x}}{x^{\frac{2}{3}}}$
I know the answer is 0. Can I simply use L'Hopitals?
You do not need L'Hopitals’ rule here. It is a lazy persons way out.
I am among those who would like to see it dropped from textbooks in basic calculus.
Note $\ln (x) = 3\ln \left( {x^{\frac{1}{3}} } \right) \le 3\left( {x^{\frac{1}{3}} } \right)$.

Therefore, $\frac{{\ln (x)}}{{x^{\frac{2}{3}} }} \le \frac{{3\left( {x^{\frac{1}{3}} } \right)}}{{x^{\frac{2}{3}} }} = \frac{3}{{x^{\frac{1}{3}} }}$ now the limit is so clear!

6. ## So plato

Originally Posted by Plato
You do not need L'Hopitals’ rule here. It is a lazy persons way out.
I am among those who would like to see it dropped from textbooks in basic calculus.
Note $\ln (x) = 3\ln \left( {x^{\frac{1}{3}} } \right) \le 3\left( {x^{\frac{1}{3}} } \right)$.

Therefore, $\frac{{\ln (x)}}{{x^{\frac{2}{3}} }} \le \frac{{3\left( {x^{\frac{1}{3}} } \right)}}{{x^{\frac{2}{3}} }} = \frac{3}{{x^{\frac{1}{3}} }}$ now the limit is so clear!
Is this sort of like how we can't use L'hopital's rule on $\lim_{n \to {\infty}}\frac{n^2}{n!}$ we know that $\forall{x}\in[4,\infty],x^2 so we know the limit is 0?