# Homework Help: Need help finding the limit of a function

Tags:
1. Apr 9, 2017

### BillyC

1. The problem statement, all variables and given/known data
Calculate limit as x approaches infinity of (e^x - x^3)

2. Relevant equations
ln e^x = x
e^(ln x) = x

3. The attempt at a solution
I tried substituting x = ln e^x and got (e^x - (ln e^x)^3). I'm pretty much lost and this is my only attempt so far.
I'm thinking that this is an indeterminate difference problem, and that I have to use L'Hospital's rule. I just can't to seem to convert this into a quotient.

2. Apr 9, 2017

### Math_QED

Hint:

$e^x - x^3 = \frac{(e^x-x^3)x^3}{x^3}$

Therefore:

$\lim_{x\to\infty} e^x - x^3 = \lim_{x\to\infty} \frac{(e^x-x^3)}{x^3} \lim_{x\to\infty} x^3$

if both limits in the right hand side exist.

3. Apr 9, 2017

### PeroK

What do you think the answer is? It's easier to prove something if you know what you are aiming for. Can you first see what the answer must be?

PS use a calculator or spreadsheet to find what happens to the function as $x$ increases.

4. Apr 9, 2017

### John Morrell

The reason this problem is tricky is that, obviously, both x^3 an e^x approach infinity as x approaches infinity. But in this case, not all infinities are born equal. Some are more infinite than others.

If you know that one of these functions is larger than the other, and remains larger, than you know that the whole function either approaches negative or positive infinity. If, on the other hand, the difference between these two functions gets smaller and smaller as x increases, then it's possible the limit converges to some finite number.

First, use this logic and some easy calculations and gut-checking to find what the limit should be, then start looking for how to prove that mathematically.

5. Apr 9, 2017

### pasmith

Can you show that $e^x - x^3 \geq 1 + x + \frac12x^2 - \frac56 x^3 + \frac1{24} x^4$ for $x \geq 0$?

6. Apr 9, 2017

### BillyC

Thank you all! This helps a lot. I suppose I just need more experience with natural logs and the number e. When I see problems involving them I'm pretty lost! Thanks again.