Limits confusion

[Loki123]

Hopefully I am in the right category. When I "solved" this problem I thought it was very easy. I got e^2014, the answer is 2015e^2014... So I lost 2014e^2014 somewhere. Where did I go wrong?

 

[Loki123]

Hopefully I am in the right category. When I "solved" this problem I thought it was very easy. I got e^2014, the answer is 2015e^2014... So I lost 2014e^2014 somewhere. Where did I go wrong?

 

[Steven G]

Why are you doing all this work? The limit equals f'(2014).

 

[Loki123]

Why are you doing all this work? The limit equals f'(2021).

What? I am supposed to get 2015e^2014

 

[Steven G]

3rd line from the bottom is wrong. Let 0, represent not 0--some number other than 0.

0/0 = 0, that is a true 0/0 = 0. However something approaching 0 divided by something approaching 0 is not necessarily 0

 

[Loki123]

3rd line from the bottom is wrong. Let 0, represent not 0--some number other than 0.

0/0 = 0, that is a true 0/0 = 0. However something approaching 0 divided by something approaching 0 is not necessarily 0

Okay, what do I do then?

 

[Steven G]

What? I am supposed to get 2015e^2014

So you are not getting f'(2014) = 2015e^2014 or did you not try? I suspect that it is the latter, so just try and you'll see that you get the correct answer.

Rewrite your limit with one change--replace x with h. Then decide if you see that the limit equals f'(2014)

 

[Steven G]

Okay, what do I do then?

Recognize that the limit is simply f'(2014). That is what the author wanted you to do with this exercise.

 

[Loki123]

So you are not getting f'(2014) = 2015e^2014 or did you not try? I suspect that it is the latter, so just try and you'll see that you get the correct answer.

Rewrite your limit with one change--replace x with h. Then decide if you see that the limit equals f'(2014)

I am having trouble understanding you. The limit's solution is supposed to be 2015e^2014, i tried to solve the problem and got e^2014.

 

[Loki123]

Recognize that the limit is simply f'(2014). That is what the author wanted you to do with this exercise.

I

So you are not getting f'(2014) = 2015e^2014 or did you not try? I suspect that it is the latter, so just try and you'll see that you get the correct answer.

Rewrite your limit with one change--replace x with h. Then decide if you see that the limit equals f'(2014)

What does f' have to do here, please explain?

 

[Deleted member 4993]

The limit equals f'(2021).

How's that?

 

[Steven G]

How's that?

Fine, the limit equals f'(2014). Are you happy now? Happy New Year!

 

[BigBeachBanana]

Have you learned l'Hopital's rule?

 

[Steven G]

I

What does f' have to do here, please explain?

To see that the limit is f'(2014) you need to do as I say. Please rewrite the limit using h instead of x. Post back that limit and we will go from there.

 

[Steven G]

Have you learned l'Hopital's rule?

The student need to recognize that this is a limit of the quotient function which is simply a derivative.

 

[Loki123]

Have you learned l'Hopital's rule?

Yess

 

[Steven G]

By definition of a derivative,

\(\displaystyle f'(a) = \lim_{h->0} \dfrac{f(a+h) - f(a)}{h}\)

Just replace a with 2014 and x with h.

 

[Loki123]

The student need to recognize that this is a limit of the quotient function which is simply a derivative.

I would like to realize that but I am not even sure I know what a quotient function is.

 

[BigBeachBanana]

Why are you doing all this work? The limit equals f'(2014).

Jomo meant that you can compute the derivative of f(x) using the product rule and evaluate it in 2014. Why do you need to do it through the definition of the derivative?

3rd line from the bottom is wrong. Let 0, represent not 0--some number other than 0.

0/0 = 0, that is a true 0/0 = 0. However something approaching 0 divided by something approaching 0 is not necessarily 0

Jomo is saying here that it's not x->inf; you should use another variable to indicate the "steps" size, typically h.

If you insist on using the definition of limit, then I'd recommend using l'Hospital's rule. First, you need to recognize why it's applicable.

 

[Steven G]

I would like to realize that but I am not even sure I know what a quotient function is.

I meant to say the difference quotient. It is the quotient used in the definition of a derivative.
Seriously, instead of just saying that you don't see what I am saying or that you don't understand what I am saying why not take my advice and rewrite the limit with two minor changes--replace 2014 with a and x with h. Then it will look exactly like the derivative formula. You need to simple practice noticing the pattern.