The problem:

Find the limit as h→0 for (e^(2+h)-e^2) / h

I really have no idea how to do this.

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- May 1st 2010, 11:19 AMRangerKimmyFinding a limit?
The problem:

Find the limit as h→0 for (e^(2+h)-e^2) / h

I really have no idea how to do this. - May 1st 2010, 11:24 AMFailure
- May 1st 2010, 11:24 AMharish21
As h->0, you get $\displaystyle \frac{0}{0}$

Use L'Hopital's rule

The derivative of the numerator is:

$\displaystyle \frac{d}{dh} (e^{2+h} -e^2) = \frac{d}{dh} (e^2 \times e^h - e^2) = e^2.e^h-0=e^{2+h}$

and the derivative of the denominator is $\displaystyle \frac{d}{dh} h = 1$

by L'hopital's rule , you then have:

$\displaystyle \lim_{h->0} \frac{e^{2+h}}{1}$ - May 1st 2010, 11:34 AMRangerKimmy
- May 1st 2010, 11:38 AMFailure
Well it's the slope of the tangent of the curve $\displaystyle y=e^x$ at the point $\displaystyle (0,1)$, that is: the limit of the slope of the secant with one fixed point in $\displaystyle (0,e^0)=(0,1)$ and the other point moving towards that fixed point ($\displaystyle h\to 0$).

Yes, this limit has the value 1. - May 1st 2010, 11:41 AMRangerKimmy