Derivatives & Limits: Solving "Does Not Exist

[MitsuShai]

Homework Statement

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

The Attempt at a Solution

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

 

[Mark44]

Homework Statement

1. Find the differential of the function (dy)
y=(x^4−7)^7

2. Evaluate.
lim (x+1)/(x^2+4x+3)
x-->3+


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

The Attempt at a Solution

1. I typed this in 7(x^4-7)^6 * 4x^3 and it was wrong, so I typed this in 28x^3(x^4-7)^6 and it was also wrong.

You are missing dx. The rest is correct.

2. I got 1/6, and it was wrong. (I factored it out and canceled the common factors)

I get 1/6 for the limit as well.

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

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.

People are telling me my answers are right, but there is no away there could be this many errors on the homework...I really can't figure out what I am doing wrong.

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.

 

[MitsuShai]

http://s324.photobucket.com/albums/k327/ProtoGirlEXE/?action=view&current=100_0527.jpg

 

[The Chaz]

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

 

[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.

 

[MitsuShai]

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

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

 

[Mark44]

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+)?

 

[MitsuShai]

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+)?

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...

 

[Mark44]

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?