# Thread: Limit where x is a power

1. ## Limit where x is a power

So I have the answer to this problem, and I know it's correct, but I want to be sure that the moves I made were legal to get there. Here's the problem:

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

Here's what I did:

$\displaystyle =\frac{lim_{x\to\infty}e^{x^2}}{lim_{x\to\infty}e^ x}$$\displaystyle =\frac{e^{lim_{x\to\infty}x^2}}{e^{lim_{x\to\infty }x}}$$\displaystyle =\frac{e^1}{e^1}=1$

Were all my moves legal? Specifically, moving the limit sign up into the exponent. Is there a rule that let's me say $\displaystyle \lim_{x \to a}B^x=B^{\lim_{x \to a}x}$ ?

2. ## Re: Limit where x is a power

Originally Posted by beebe
So I have the answer to this problem, and I know it's correct, but I want to be sure that the moves I made were legal to get there. Here's the problem:

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

Here's what I did:

$\displaystyle =\frac{lim_{x\to\infty}e^{x^2}}{lim_{x\to\infty}e^ x}$$\displaystyle =\frac{e^{lim_{x\to\infty}x^2}}{e^{lim_{x\to\infty }x}}$$\displaystyle =\frac{e^1}{e^1}=1$

Were all my moves legal? Specifically, moving the limit sign up into the exponent. Is there a rule that let's me say $\displaystyle \lim_{x \to a}B^x=B^{\lim_{x \to a}x}$ ?
The correct answer is $\displaystyle \infty$.

How did you get that $\displaystyle \lim_{x\to\infty}x^2=\lim_{x\to\infty}x=1$?

3. ## Re: Limit where x is a power

Sorry, I combined multiple problems when I was typing. The infinity symbols should be 1's. Let me try again:

$\displaystyle \lim_{x \to 1}e^{x^2-x}$$\displaystyle =\frac{lim_{x \to 1}e^{x^2}}{lim_{x \to 1}e^x}$$\displaystyle =\frac{e^{lim_{x \to 1}x^2}}{e^{lim_{x \to 1}x}}$$\displaystyle =\frac{e^1}{e^1}=1 4. ## Re: Limit where x is a power Originally Posted by beebe Sorry, I combined multiple problems when I was typing. The infinity symbols should be 1's. Let me try again: \displaystyle \lim_{x \to 1}e^{x^2-x}$$\displaystyle =\frac{lim_{x \to 1}e^{x^2}}{lim_{x \to 1}e^x}$$\displaystyle =\frac{e^{lim_{x \to 1}x^2}}{e^{lim_{x \to 1}x}}$$\displaystyle =\frac{e^1}{e^1}=1$
That's right.

An easier way would be $\displaystyle \lim_{x \to 1}e^{x^2-x}=e^{\lim_{x \to 1}x^2-x}=e^0=1$

5. ## Re: Limit where x is a power

Originally Posted by beebe
Sorry, I combined multiple problems when I was typing. The infinity symbols should be 1's. Let me try again:
$\displaystyle \lim_{x \to 1}e^{x^2-x}$$\displaystyle =\frac{lim_{x \to 1}e^{x^2}}{lim_{x \to 1}e^x}$$\displaystyle =\frac{e^{lim_{x \to 1}x^2}}{e^{lim_{x \to 1}x}}$$\displaystyle =\frac{e^1}{e^1}=1$
Well you chanced the question!
You did not have to do that much because $\displaystyle e^0=1$