# Math Help - Limit Problems

1. ## Limit Problems

These are the only two i couldn't figure out. If someone could help, that'd be great. Here they are:

1) Limit where h approaches 0- {Square root(4 + h) - 2}/h

In words, the limit where h approaches zero of the square root of four plus h minus 2 all over h.

2) Limit where h approaches 0- 1/{square root(4 + h)} - 1/{2}/h

In words, the limit where h approaches zero of 1 divided by the square root of four plus h minus one half all over h.

Please help, need by tomorrow morning! Thanks!

2. 1) lim ((squ(4+h)-2)/h)((squ(4+h)+2)/(squ(4+h)+2) =

lim (4+h-4)/(h*squ(4+h)+2) =

lim 1/(squ(4+h)+2 =

1/(2+2) =

1/4

2)

0

I don't know if i understood part 2 correctly...

3. Ok, well thank you for the first one.

4. Originally Posted by Blue Griffin
2) Limit where h approaches 0- 1/{square root(4 + h)} - 1/{2}/h
well, what are your options here? try to simplify the top so that we cancel the h in the bottom, that way we can take the limit. how do we simplify the top? well, we can start by combining the fractions.

$\lim_{h \to 0} \frac {\frac 1{\sqrt {4 + h}} - \frac 12}h$

$= \lim_{h \to 0} \frac {\frac {2 - \sqrt{4 + h}}{2 \sqrt{4 + h}}}{h}$

$\lim_{h \to 0} \frac {2 - \sqrt{4 + h}}{2h \sqrt {4 + h}}$

well, now what? the h doesn't seem to cancel, so now we apply a common trick, conjugation of the numerator...

$\Rightarrow \lim_{h \to 0} \frac {2 - \sqrt{4 + h}}{2h \sqrt {4 + h}} \cdot \frac {2 + \sqrt{4 + h}}{2 + \sqrt {4 + h}}$

now continue and see where you get with this. your answer should be $- \frac 1{16}$

5. Yeah definitely didn't read the second one correctly

6. Thanks a lot guys, i understand it now, and i finished my homework!