# Thread: Limits and Improper Integrals

1. ## Limits and Improper Integrals

I have a test tomorrow over L'Hopital's Rule and Improper Integrals. I understand how they work, but I'm having trouble figuring out when an integral (or it's limit I should say) is going toward infinity or towards 0. For example, how do I know what the limit as x approaches infinity of e^x is? And the whole infinity/zero or zero/infinity what those approach... I just don't understand the way the zero or infinity thing works with limits. Any sort of explanation/example would be helpful to me at this point.

2. Are you asking for limits at infinity?

Like $\displaystyle \lim_{x\to\infty} e^x$?

3. ## Yes

Originally Posted by wingless
Are you asking for limits at infinity?

Like $\displaystyle \lim_{x\to\infty} e^x$?

Yes, pretty much. How can a limit be infinite? I guess it just seems like an oxymoron to me and I can't coherce my brain into grasping it...

Thanks!

4. You have to think like this, what will happen to this function when x grows and grows..

If you know how to find normal limits, we can start with some examples and I'm sure you'll get what it is like.

Firstly, $\displaystyle \lim_{x\to\infty} \frac{1}{x} = 0$.

Exponential:
$\displaystyle \lim_{x\to\infty} e^x$
For $\displaystyle x=1$, it's $\displaystyle e = 2.7182$
For $\displaystyle x=2$, it's $\displaystyle e^2 = 7.3891$
For $\displaystyle x=5$, it's $\displaystyle e^5 = 148.41$
For $\displaystyle x=100$, it's $\displaystyle e^100 = 2.688\cdot 10^{43}$
For $\displaystyle x=100000$, it's $\displaystyle e^100000 = 2.806\cdot 10^{43429}$
...
When x grows and grows and approaches infinity, $\displaystyle e^x$ will grow too, and will also approach infinity.

As a general rule,
If $\displaystyle 0<r<1$, $\displaystyle \lim_{x\to\infty} r^x = 0$

If $\displaystyle r>1$, $\displaystyle \lim_{x\to\infty} r^x = \infty$

You can solve limits approaching negative infinity using,
$\displaystyle r^{-\infty} = \frac{1}{r^{\infty}}$

--------------

Polynomial Fractions:

$\displaystyle \frac{x^3-2x^2+5}{x^2+2x-2}$
General method: Find the degree of the polynomial (degree of the term that has the highest degree) of the denominator. For example if that degree is 3, divide both numerator and denominator by $\displaystyle x^3$.

$\displaystyle \frac{x^3-2x^2+5}{x^2+2x-2}$

Highest degree of $\displaystyle x^2+2x-2$ is 2. Divide both numerator and denominator by $\displaystyle x^2$.

$\displaystyle \frac{\frac{x^3-2x^2+5}{x^2}}{\frac{x^2+2x-2}{x^2}}$

$\displaystyle \frac{\frac{x^3}{x^2} - \frac{2x^2}{x^2} + \frac{5}{x^2}}{\frac{x^2}{x^2} + \frac{2x}{x^2}-\frac{2}{x^2}}$

$\displaystyle \frac{x - 2 + \not\frac{5}{x^2}}{1 + \not\frac{2}{x}-\not\frac{2}{x^2}}$

$\displaystyle x-2$. This will approach $\displaystyle \infty$.

$\displaystyle \lim_{x\to\infty}\frac{3x^2 -2x + 3}{2x^2 - x + 1}$

Highest degree of $\displaystyle 2x^2 - x + 1$ is 2, so divide both numerator and denominator by $\displaystyle x^2$

$\displaystyle \frac{3 - \not\frac{2}{x} + \not\frac{3}{x^2}}{2 - \not\frac{1}{x} + \not\frac{1}{x^2}}$

$\displaystyle \frac{3}{2}$. This is the limit.

General method:
If the degree of numerator is greater than denominators, the limit will approach $\displaystyle \pm \infty$.
If the degree of numerator is less than denominators, the limit will approach $\displaystyle 0$.
If the degree of numerator is equal to denominators, the limit is the ratio of the coefficients of the terms with highest degrees of numerator and denominator.

5. ## Okay

Okay, so think about x/infinity as dividing x by a really really really really big number, so it approaches zero... and infinity/x would be infinity because no matter how big x gets, infinity is just infinity so it "wins out" over the x... I guess the same would be true for zero etc. Thanks ALOT for your help, wingless, that was an awesome post.

6. Thanks ! I'm really glad that helped you.

Here are a few more things about it:

Comparing terms is another great method for finding limits.

Let's have $\displaystyle \lim_{x\to\infty}x^3-x^2$. When you put $\displaystyle \infty$, it gives $\displaystyle \infty - \infty$. WARNING: $\displaystyle \infty - \infty$ is NOT ZERO! It's called an indeterminate form. You could apply l'hopital's rule but that would be slicing an apple with a chainsaw.

If you have a polynomial (when x approaches infinity), the term with the highest degree will EAT the others. In our example, $\displaystyle \lim_{x\to\infty}x^3-x^2$, $\displaystyle x^3$ has the highest degree and it'll EAT the others. When $\displaystyle x^3$ and $\displaystyle x^2$ approaches infinity, $\displaystyle x^3$ will approach much more faster than $\displaystyle x^2$ and this'll make $\displaystyle x^2$ insignificant. So we can throw $\displaystyle x^2$ away.

$\displaystyle \lim_{x\to\infty}x^3-x^2$

$\displaystyle \lim_{x\to\infty}x^3-\not x^2$

$\displaystyle \lim_{x\to\infty}x^3 = \infty$.

Note: Exponentials ALWAYS grow faster than ANY x term. $\displaystyle 1.1^{x}$ will grow faster than even $\displaystyle 10000000 x^{1000000}$. Of course $\displaystyle 0.9^x$ will approach 0, I was talking about $\displaystyle r^x$ with $\displaystyle r>1$.

So,

$\displaystyle \lim_{x\to\infty} e^x - x^{100}$

$\displaystyle \lim_{x\to\infty} e^x - \not x^{100}$

$\displaystyle \lim_{x\to\infty} e^x = \infty$

$\displaystyle \lim_{x\to\infty}x^5 - x^4 + 10x^3 - x + 5$

$\displaystyle \lim_{x\to\infty}x^5 - \not x^4 + \not 10 \not x^3 - \not x + \not 5$

$\displaystyle \lim_{x\to\infty}x^5 = \infty$

We can use this on fractions.

$\displaystyle \lim_{x\to\infty}\frac{-4x^2 +2x-10}{3x^2 + x - 5}$

$\displaystyle \frac{-4x^2 +\not {2x}-\not 10}{3x^2 + \not x - \not 5}$

$\displaystyle \frac{-4x^2}{3x^2} = -\frac{4}{3}$

$\displaystyle \lim_{x\to\infty} \frac{2^x + 5x^2 - 3x + 2}{3^x - 6x + 2}$

$\displaystyle \frac{2^x + \not 5x^2 - \not 3x + \not 2}{3^x - \not 6x + \not 2}$

$\displaystyle \frac{2^x}{3^x}$

$\displaystyle {\left ( \frac{2}{3}\right ) }^x$

As $\displaystyle 0<\frac{2}{3}<1$, this limit is $\displaystyle 0$.