Evaluating a Limit

[SunnySummerTime]

lim [(√9+h)-3]
h->0

The problem states evaluate the limit. If it exists. I'm pretty sure that I should use the conjugate to solve this but after multiple tries I can't seem to get it. I know the answer is 1/6 but I'm not sure how to get there. Oh, and the way it's typed doesn't really show it but the square root is of 9+h not just the 9. Any help on this would be greatly appreciated! Thanks!

 

[JeffM]

lim [(√9+h)-3]
h->0

The problem states evaluate the limit. If it exists. I'm pretty sure that I should use the conjugate to solve this but after multiple tries I can't seem to get it. I know the answer is 1/6 but I'm not sure how to get there. Oh, and the way it's typed doesn't really show it but the square root is of 9+h not just the 9. Any help on this would be greatly appreciated! Thanks!

Are you supposed to find the limit and demonstrate its validity using the delta-epsilon method? Or are you supposed to use the basic laws of limits?

And what EXACTLY is the statement of the problem, and why do you say the answer is 1/6?

\(\displaystyle \sqrt{9 + .1} - 3 \approx 3.02 - 3 = .02.\)

\(\displaystyle \sqrt{9 + .01} - 3 \approx 3.002 - 3 = .002.\)

\(\displaystyle \sqrt{9 + .001} - 3 \approx 3.0002 - 3 = .0002.\)

Does that look as though it is approaching 0.166....

 

[srmichael]

lim [(√9+h)-3]
h->0

The problem states evaluate the limit. If it exists. I'm pretty sure that I should use the conjugate to solve this but after multiple tries I can't seem to get it. I know the answer is 1/6 but I'm not sure how to get there. Oh, and the way it's typed doesn't really show it but the square root is of 9+h not just the 9. Any help on this would be greatly appreciated! Thanks!

Are you sure the problem isn't:

\(\displaystyle \lim_{h\rightarrow0}\frac{\sqrt{9+h}-3}{h}\)

 

[HallsofIvy]

Assuming that it is actually \(\displaystyle \frac{\sqrt{9+h}- 3}{h}\), as Sir Michael suggests, what do you get when you multiply both numerator and denominator by \(\displaystyle \sqrt{9+h}+ 3\)?