# Interpreting Limits

• Aug 5th 2009, 08:05 AM
skeske1234
Interpreting Limits
Evaluate the limit of the difference quotient. Interpret the limit as the slope of the tangent to a curve at a specific point.

lim [(2+h)^2-4]/h
h->0

my attempt:
I need help with the secondi part. I found the derivative which in this case is M_t=4 to the graph y=x^2 at x=_______? I am not sure how to find x =? at the specific point.

I tried to put M_t=4 into the original eqtn and solve for h but that didn't work. Please provide detailed steps to finishing this problem. Thank you!
• Aug 5th 2009, 08:16 AM
skeeter
Quote:

Originally Posted by skeske1234
Evaluate the limit of the difference quotient. Interpret the limit as the slope of the tangent to a curve at a specific point.

lim [(2+h)^2-4]/h
h->0

my attempt:
I need help with the secondi part. I found the derivative which in this case is M_t=4 to the graph y=x^2 at x=_______? I am not sure how to find x =? at the specific point.

I tried to put M_t=4 into the original eqtn and solve for h but that didn't work. Please provide detailed steps to finishing this problem. Thank you!

$f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$

for your problem, note that $f(x) = x^2$

the limit is $f'(2)$
• Aug 5th 2009, 08:17 AM
Plato
$\frac{{\left( {2 + h} \right)^2 - 4}}
{h} = \frac{{\left( {4 + 4h + h^2 } \right) - 4}}
{h} = 4 + h,\,h \ne 0$
• Aug 5th 2009, 08:18 AM
skeske1234
Quote:

Originally Posted by Plato
$\frac{{\left( {2 + h} \right)^2 - 4}}
{h} = \frac{{\left( {4 + 4h + h^2 } \right) - 4}}
{h} = 4 + h,\,h \ne 0$

the answer is: 4, slope of tangent to y=x^2 at x=2

My question is: how do I get x=2?
• Aug 5th 2009, 08:25 AM
Plato
Quote:

Originally Posted by skeske1234
the answer is: 4, slope of tangent to y=x^2 at x=2
My question is: how do I get x=2?

That question really do not make any sense. Does it?

If the question is, "find the x value where $y=x^2$ has slope 4."
You would look at the derivative: $2x=4$.
From which it i clear that $x=2$.
• Aug 5th 2009, 08:25 AM
lancekam
So you have already found the slope, m=4. I think you want the tangent line now. Based on your work, you used the difference quotient and pugged x=2 in to find the slope at this x value. So, at x=2 the slope is 4. So you have an x value, you now need a y-value in order to talk about the tangent line. Well, you know the tangent line has to touch our orinignal graph, $y=x^2$, so to find the y-value, just plug in our x-value. If x=2, y=4. So the point our line in tangent to is (2,4) with a slope of 4. Using point-slope form, our tangent line is:
$y-4=4(x-2)$
or
$y=4x-4$