1. ## Derivatives and limits

Hi,

I hope someone can help. I want to know whether my thinking is right or not about the following question:

For 15a - Not exactly sure how to interpreted this one and would really appreciate help on this.

For 15 b - I interpreted this limit as the derivative for e^x at x = 2, which implies that the limit for 15b is equal to e^2.

- Olivia

2. ## Re: Derivatives and limits

Oh I think I get 15a now! I believe that the limit represents the derivative for e^x at x = 0! The answer for that limit is 1 since the e^0 = 1. Yay

3. ## Re: Derivatives and limits

$\dfrac {d}{dx} e^x =\displaystyle \lim_{h\to 0} \dfrac{ e^{x+h} - e^x}{h}$

$\left(\left . \dfrac{d}{dx} e^x \right |_{x=0} \right)= \displaystyle \lim_{h\to 0} \dfrac{ e^h -1}{h}$

$\dfrac{d}{dx}e^x = e^x$

$\left . e^x \right |_{x=0} = 1$

$\displaystyle \lim_{h\to 0} \dfrac{ e^h -1}{h}=1$

see if you can do the 2nd one.

4. ## Re: Derivatives and limits

Originally Posted by otownsend
Oh I think I get 15a now! I believe that the limit represents the derivative for e^x at x = 0! The answer for that limit is 1 since the e^0 = 1. Yay
ok.. let's look at 2nd one then :P

same deal but $x=2$

5. ## Re: Derivatives and limits

No.

I think it's $\displaystyle \left.\frac{\mathrm d}{\mathrm dx} e^{2+x}\right|_{x=0}$

6. ## Re: Derivatives and limits

Yeah thanks! makes sense.

8. ## Re: Derivatives and limits

Originally Posted by otownsend
Hi,

I hope someone can help. I want to know whether my thinking is right or not about the following question:

For 15a - Not exactly sure how to interpreted this one and would really appreciate help on this.

For 15 b - I interpreted this limit as the derivative for e^x at x = 2, which implies that the limit for 15b is equal to e^2.

- Olivia
Actually in its actual form, the solution does not need to get derivative of the given forms. What it all asks is to put the limits i.e. $h \to 0$. But if you put those limits in place, you will soon notice that the term will become $\frac{0}{0}$ which is an indeterminate form.
Hence then you are bound to use the L'Hospital rule which means you have to derivate the Numerator and Denominator to get the proper answer. Hope this makes sense.

9. ## Re: Derivatives and limits

That's a circular argument. There's no need, no any real use for l'Hôpital here.

10. ## Re: Derivatives and limits

Originally Posted by otownsend
Oh I think I get 15a now! I believe that the limit represents the derivative for e^x at x = 0! The answer for that limit is 1 since the e^0 = 1.
That absolutely correct.

Now for 15b, note that $\dfrac{e^{2+h}-e^2}{h}=e^2\cdot\dfrac{e^{h}-1}{h}$, hence we can use part a to answer part b.