Finding a limit?

**RangerKimmy** · May 1st 2010, 11:19 AM

The problem:

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

I really have no idea how to do this.

**Failure** · May 1st 2010, 11:24 AM

Originally Posted by RangerKimmy

The problem:

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

I really have no idea how to do this.

Consider this:
$\lim_{h\to 0}\frac{e^{2+h}-e^2}{h}=\lim_{h\to 0}\frac{e^2(e^h-1)}{h}=e^2\cdot\lim_{h\to0}\frac{e^h-1}{h}$

Have you seen the remaining limit $\lim_{h\to0}\frac{e^h-1}{h}=\lim_{h\to0}\frac{e^h-e^0}{h}$ before?

**harish21** · May 1st 2010, 11:24 AM

Originally Posted by RangerKimmy

The problem:

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

I really have no idea how to do this.

As h->0, you get $\frac{0}{0}$

Use L'Hopital's rule

The derivative of the numerator is:

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

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

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

$\lim_{h->0} \frac{e^{2+h}}{1}$

**RangerKimmy** · May 1st 2010, 11:34 AM

Originally Posted by Failure

Consider this:
$\lim_{h\to 0}\frac{e^{2+h}-e^2}{h}=\lim_{h\to 0}\frac{e^2(e^h-1)}{h}=e^2\cdot\lim_{h\to0}\frac{e^h-1}{h}$

Have you seen the remaining limit $\lim_{h\to0}\frac{e^h-1}{h}=\lim_{h\to0}\frac{e^h-e^0}{h}$ before?

I follow what you have done.
However, if that limit is something I should already know, then no, I haven't seen it before. Does it equal 1?

**Failure** · May 1st 2010, 11:38 AM

Originally Posted by RangerKimmy

I follow what you have done.
However, if that limit is something I should already know, then no, I haven't seen it before. Does it equal 1?

Well it's the slope of the tangent of the curve $y=e^x$ at the point $(0,1)$ , that is: the limit of the slope of the secant with one fixed point in $(0,e^0)=(0,1)$ and the other point moving towards that fixed point ( $h\to 0$ ).
Yes, this limit has the value 1.

**RangerKimmy** · May 1st 2010, 11:41 AM

Originally Posted by Failure

Well it's the slope of the tangent of the curve $y=e^x$ at the point $(0,1)$ , that is: the limit of the slope of the secant with one fixed point in $(0,e^0)=(0,1)$ and the other point moving towards that fixed point ( $h\to 0$ ).
Yes, this limit has the value 1.

Ohhh. Okay, thanks! This has helped tremendously!

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