1. ## Limit Problem

Hi folks. I'm stuck on evaluating a limit. Any help would be appreciated.

$\lim_{x\to\infty}\frac{x}{\sqrt{x^2+1}}$

Since the limit produces the form $\frac{\infty}{\infty}$, applying L'Hopital's rule gives the reciprocal function, which in turn is of the form $\frac{\infty}{\infty}$. Applying L'Hopital's rule again gives the original function. Help!

2. Originally Posted by kaiser0792
Hi folks. I'm stuck on evaluating a limit. Any help would be appreciated.

$\lim_{x\to\infty}\frac{x}{\sqrt{x^2+1}}$

Since the limit produces the form $\frac{\infty}{\infty}$, applying L'Hopital's rule gives the reciprocal function, which in turn is of the form $\frac{\infty}{\infty}$. Applying L'Hopital's rule again gives the original function. Help!
This is really just intuitive. Sub in a really big value for x and you can see that the limit has to be 1.

3. This is true, thank you. While on the subject, is $\infty^0$ equal to 1 (b/c any number to the zero power is equal to 1)?

4. Originally Posted by kaiser0792
This is true, thank you. While on the subject, is $\infty^0$ equal to 1 (b/c any number to the zero power is equal to 1)?
Interesting question - I would say it is.
BTW 0^0 is not equal to 1 (it is indeterminate).

5. Originally Posted by kaiser0792
This is true, thank you. While on the subject, is $\infty^0$ equal to 1 (b/c any number to the zero power is equal to 1)?
You have the classic problem and the answer is...the question doesn't make sense.

Think about. This notion you have of the function $f(x,y)=x^y$ is only defined in terms of real numbers. Is $\infty\in\mathbb{R}$? No, so it's not a debate of "is infinity so big it makes the zeroth power go away or is the zero so small it makes the infinity go away" the answer is you really can't ask the question without introducing some new number $\infty$ and doing this (at least in th obvious way) screws up the algebraic structure of $\mathbb{R}$ ( $\infty+1=\infty\implies 1=0$ and since this isn't true we can't additively cancel things). So, if you're asking what is $\lim_{x\to 0}f(x)^{g(x)}$ where $f\to\infty,g\to 0$ the answer is...it can be any real number you want.

6. I knew that, thanks for the help. Greatly appreciated!

7. Originally Posted by Drexel28
You have the classic problem and the answer is...the question doesn't make sense.

Think about. This notion you have of the function $f(x,y)=x^y$ is only defined in terms of real numbers. Is $\infty\in\mathbb{R}$? No, so it's not a debate of "is infinity so big it makes the zeroth power go away or is the zero so small it makes the infinity go away" the answer is you really can't ask the question without introducing some new number $\infty$ and doing this (at least in th obvious way) screws up the algebraic structure of $\mathbb{R}$ ( $\infty+1=\infty\implies 1=0$ and since this isn't true we can't additively cancel things). So, if you're asking what is $\lim_{x\to 0}f(x)^{g(x)}$ where $f\to\infty,g\to 0$ the answer is...it can be any real number you want.
"only defined in terms of real numbers" ???
What about eg i^x ? I realise this is a different question entirely, but powers of complex numbers are defined.

8. I understand, my problem was defined in terms of $\mathbb{R}$. Thank you!

9. Originally Posted by Debsta
"only defined in terms of real numbers" ???
What about eg i^x ? I realise this is a different question entirely, but powers of complex numbers are defined.
Firstly, the difference in subtetly of the definitions is astounding. If you have any knowledge of complex analysis you should know this well considering for example the difference between $\text{Ln}(z)$ and $\ln(z)$. Secondly, there is a fine difference. The point here (that I am concerned with) is an algebraic one. As it stands $\mathbb{R}$ is a complete ordered field and in fact the only one up to isomorphism and so unless you want to completely redefine many aspects of our number system which we may take for granted you'll have to define $\mathbb{R}\cup\{\infty\}$ in a way which makes it essentially the same as $\mathbb{R}$.

For example as I said you always here $\infty+1=\infty$ but then $\mathbb{R},+$ wouldn't even be a group.

Now, you can append the point " $\infty$" for topological purposes (this is the Alexandroff compactification of locally compact Hausdorff spaces) but you'll get the unit circle $\mathbb{S}^1$

I think I'm rambling now...sorry it's late.

10. Originally Posted by Debsta
Interesting question - I would say it is.
BTW 0^0 is not equal to 1 (it is indeterminate).
At first glance I thought this true as well...

But consider the functions...

$f(x) = x^x$ and $g(x) = \frac{1}{x}$

Then,

$\lim_{x \to \infty} f(x) = \infty$

and

$\lim_{x \to \infty} g(x) = 0$

But...

$\lim_{x \to \infty} f(x)^{g(x)} = \infty$...

In fact,

$f(x)^{g(x)} = x$ so we can quite easily see it diverges.

11. Originally Posted by kaiser0792
Hi folks. I'm stuck on evaluating a limit. Any help would be appreciated.

$\lim_{x\to\infty}\frac{x}{\sqrt{x^2+1}}$

Since the limit produces the form $\frac{\infty}{\infty}$, applying L'Hopital's rule gives the reciprocal function, which in turn is of the form $\frac{\infty}{\infty}$. Applying L'Hopital's rule again gives the original function. Help!
Algebraic manipulation is your friend here.

Divide top and bottom by the square root of the highest power.

In other words, multiply top and bottom by $\frac{1}{\sqrt{x^2}} = \frac{1}{x}$.

So $\lim_{x \to \infty}\frac{x}{\sqrt{x^2 + 1}} = \lim_{x \to \infty}\frac{x\left(\frac{1}{\sqrt{x^2}}\right)}{\ sqrt{x^2 + 1}\left(\frac{1}{\sqrt{x^2}}\right)}$

$= \lim_{x \to \infty}\frac{1}{\sqrt{1 + \frac{1}{x^2}}}$

$= \frac{1}{\sqrt{1}}$

$= 1$.

At first glance I thought this true as well...

But consider the functions...

$f(x) = x^x$ and $g(x) = \frac{1}{x}$

Then,

$\lim_{x \to \infty} f(x) = \infty$

and

$\lim_{x \to \infty} g(x) = 0$

But...

$\lim_{x \to \infty} f(x)^{g(x)} = \infty$...

In fact,

$f(x)^{g(x)} = x$ so we can quite easily see it diverges.
As I said, you can make it any real number.

$\lim_{x\to0}\left(e^{\frac{\ln(a)}{|x|}}\right)^{| x|}=e^{\ln(a)}=a$

13. "ProveIt" your answer makes the most sense to me. Not to take away from everyone else's explanations. The algebraic manipulation is what I was attempting. I just didn't try your approach. Thank you very much!