Question

lim as h--->0 (find in terms of constant a)
sqrt(3(a+h))-sqrt(3a) / h
I know I have to multiply by the conjugate, but how do I cancel the bottom?? please help!!!

Answer 1

\[\frac{ \sqrt{3a+h}-\sqrt{3a} }{ h }\]

Answer 2

is that what you mean?

Answer 3

yes :)

Answer 4

as h--->0

Answer 5

ok just multiply by the conjugate like you said first so multiply by
\[\sqrt{3a+h}-\sqrt{3a}\]

Answer 6

\[\sqrt{3a+h}+\sqrt{3a}\]

Answer 7

sorry plus 3a not minus

Answer 8

wait you mean + \[\sqrt{3a}\]

Answer 9

yeah thanks LG2

Answer 10

np:) just go on..

Answer 11

ok I did that and then I cancelled out the squares with the square roots and I was left with (3h) / \[\sqrt{3(a+h)}\] +\[\sqrt{3a}\]

Answer 12

I'm not sure what to do from there :/

Answer 13

you might have to retype that after I multiply by the conjugate I have \[\frac{ 3a+3h-3a }{ h \sqrt{3(a+h}+\sqrt{3a} }\]

Answer 14

the 3as cancel

Answer 15

yayyy I got that right :D but how do the square roots on the bottom cancel?:O

Answer 16

you cancel out 1 h so you have\[\frac{ 2h }{ \sqrt{3(a+h)}+\sqrt{3a} }\]

Answer 17

you just have to plug in h and then rationalize the denominator

Answer 18

so you really have\[\frac{ 2h }{ \sqrt{3a}+\sqrt{3a} }\]

Answer 19

or well zero on top

Answer 20

so the limit is 0

Answer 21

most likely more exactly the limit is\[\frac{ 0 }{ 2\sqrt{3a} }\] since we dont know what a is

Answer 22

and simplifying the denominator we are left with\[\frac{ 0 }{ 6a}\]

Answer 23

ohh ok so the h on the bottom became 0 right so that's why you had \[\sqrt{3a}\] + \[\sqrt{3a}\]

Answer 24

wait how did you get 6a??

Answer 25

yeah it became zero because we used direct substitution

Answer 26

we multiplied \[2\sqrt{3a}*\sqrt{3a}\] we were left with 3a*2

Answer 27

which is 6a

Answer 28

ohh ok so you multiplied to get rid of the square root?

Answer 29

uhm just a little correction: