1. ## Evaluating Limits

i'm so lost on this problem.. I don't even know where to start on this one. Help me guys

Evaluate.

Lim (3+h)^-1 - 3^-1 / h
h->0

2. $\lim_{h \to 0} \frac{1}{3 + h} - 3^{-\frac{1}{h}}$

Does it look like this?

3. 3 ways:
- Combine fractions in the numerator
- Multiply by 3(3+h) on both sides of the fraction
- Treat it as a derivative and use power/quotient rule to evaluate it at h = 0.

4. Originally Posted by 11rdc11
$\lim_{h \to 0} \frac{1}{3 + h} - 3^{-\frac{1}{h}}$

Does it look like this?
It looks like he meant:
$\lim_{h \to 0} \frac{\displaystyle\frac{1}{3+h} - \frac{1}{3}}{h}$

I could be wrong though.

5. Originally Posted by Chop Suey
It looks like he meant:
$\lim_{h \to 0} \frac{\displaystyle\frac{1}{3+h} - \frac{1}{3}}{h}$

I could be wrong though.
Hopefully becuase the other limit a little bit harder to do.

6. Hey guys,

Sorry to bump this thread up again but i've been trying to figure the answer out and keep getting 0/9, which is incorrect. Any help on pointing me in the right direction? Thanks everybody.

7. $\lim_{h \to 0} \frac{\displaystyle\frac{1}{3+h} - \frac{1}{3}}{h}$

8. $\lim_{h \to 0} \frac{\displaystyle\frac{1}{3+h} - \frac{1}{3}}{h}$

ok like chop suey said find a common denominator

$\lim_{h \to 0}\frac{\frac{-h}{9+3h}}{h}$

$\lim_{h \to 0} \frac{-h}{h(9+3h)}$

Can you finish up from here?

9. the common denominator would be (3+h)(3) right? So I multiply that top and bottom and I get this

3 - 3 - h / h (3+h)(3)

am I doing something wrong?

10. but if you factored out a 3 wouldn't it be 2 + h ?

11. Yep your right I made a mistake. Thanks for pointing that out. The right should be $-\frac{1}{9}$

12. Thanks alot man you really helped me understand the concept .

13. Sorry to bump this thread again but I had another question pertaining to the problem from before.

The slope would equal -1/9 right? My professor had this answer down.

"The slope of the tangent line to y = 1/x at the point (3, 1/3) is -1/9."

How did he get the points (3,1/3) and y = 1/x?

14. The derivative is the slope of the tangent line at a point x=a. The definition is:
$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$

This is very similar to your limit. The given question had already plugged in the point, but one can find the original function easily. Let's revert 3 back to x:

$f(x+h) = \frac{1}{x+h}$

$f(x) = \frac{1}{x}$

$f(3) = \frac{1}{3}$

So $-\frac{1}{9}$ is the slope of the tangent line to point $(3, \frac{1}{3})$

15. So basically f(x) = 1/x was the original equation? All I had to do was plug the X value which is 3 back into the original equation to find my Y value?

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