# Derivatives & Limits

1. May 2, 2010

### MitsuShai

1. The problem statement, all variables and given/known data

lim (e^(7x)-1)/x^2
x-->0

3. The attempt at a solution

I typed in "does not exist" and it was wrong.

Last edited: May 2, 2010
2. May 2, 2010

### Staff: Mentor

You are missing dx. The rest is correct.
I get 1/6 for the limit as well.
I get the same. For this problem, the right-hand limit is infinity and the left-hand limit is -infinity, so the two-sided limit does not exist.
I don't either. All I can suggest is to make sure the problems you posted here are the same ones that are in your book or wherever you got them.

3. May 2, 2010

### MitsuShai

Last edited: May 2, 2010
4. May 2, 2010

### The Chaz

2. I suspected as much! It's the limit as x approaches NEGATIVE three.
3. Use L'Hopital's rule...once

Last edited: May 2, 2010
5. May 2, 2010

### tangibleLime

Concerning #3,

Since $$\lim_{x\rightarrow 0} \frac{e^(7x)-1}{x^2} = \frac{0}{0}$$ (called an indeterminate), you have to use L'Hopital's rule, which states that $$\lim_{x\rightarrow c} \frac{f(x)}{g(x)} = \lim_{x\rightarrow c} \frac{f'(x)}{g'(x)}$$

So, find the derivative of f(x) and g(x) and use apply the rule above.

6. May 2, 2010

### MitsuShai

wow was that a stupid mistake and for number 3 people were telling me that I couldn't use l'hopital's rule that's why I didn't.

so here's what I got
1. for number one do I just have to put dx at the end??? like this: 7(x^4-7)^6 * 4x^3 dx or like this 28x^3(x^4-7)^6 dx (it's my last chance to put the correct answer and I don't want to put the wrong thing in)
2. I factored it out and got lim 1/(x+3)
then I get 1/0..so does that mean positive infinity or something
3. this is what I did before: lim 7e^7x/2x = 7/0, so is that infinity?
Before I was thinking I should apply the rule twice because I get 7/0 and If I did apply it twice I get: lim 49e^(7x)/2= 49/2

Last edited: May 2, 2010
7. May 2, 2010

### Staff: Mentor

For #1 I would go with 28x^3(x^4 - 7)^6 dx, but the other expression is equal to this, so either should be marked as correct.

For #2, the limit is taken as x --> -3 from the right, so the limit is +infinity.

For #3, after applying L'Hopital's rule once you get 7e^(7x)/(2x) (which is what you show). Is the left side limit (x --> 0-) the same as the right side limit (x --> 0+)?

8. May 2, 2010

### MitsuShai

no, x --> 0- goes to negative infinity and x --> 0+ goes to positive infinity, so the limit does not exist, but I put "does not exist" and it was wrong....

Last edited: May 2, 2010
9. May 2, 2010

### Staff: Mentor

I don't see how that answer could be marked wrong. Can you talk to your instructor?

Are you supposed to write something like DNE?