Limits confusion

[Dr.Peterson]

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?

There are several ways this can be done, depending in part on what you have learned. You put this under algebra rather than calculus (which is where both limits and derivatives belong); but you evidently do know about both.

I like Jomo's approach of recognizing this as a difference quotient, and therefore a derivative; but L'Hopital's rule also works.

However, you can also do it directly, if you have learned that the limit of (e^x - 1)/x is 1.

But you asked where you went wrong. It's here:

You can't take x to 0 in the numerator separately from doing it in the denominator.

Keep simplifying without taking any limits, until you get to a point where you can; you'll have, as I suggested, (e^x - 1)/x.

 

[Loki123]

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?

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.

I am fairly familiar with the l'Hopital's rule and why it's applicable. I used it and managed to solve it. Thank you for reminding me of it.

 

[JeffM]

I am going to start RIGHT at the beginning. Look at post #1. YOU NEVER TOLD US WHAT THE PROBLEM WAS.

We like a lot that you showed us your work, but that still leaves us guessing what the problem is.

I am guessing that the problem was something like

Find [imath]f’(2015) \text { given that } f(x) = x * e^{2015x}.[/imath]

Are you supposed to do it from first principles? If so, as jomo has been trying patiently to tell you, your error starts on the first line.

[math]f’(x) = \lim_{h \rightarrow 0} \dfrac{f(x +h) - f(x)}{h}.[/math]
That is known as the limit of the difference quotient (or Newton quotient).

Your [math]x \rightarrow 0[/math] is entirely wrong.

Whether jono’s comments help solve the actual problem is unknowable because we do not know what the problem is.

(By the way, finding the derivative of [imath]xe^{cx}[/imath] from first principles is hard.)