Question

Hello I was wondering on how to evaluate the limit of:
sqrt(6+h)-sqrt(6)
--------------
h
Lim
h->0

Answer 1

multiply top and bottom by sqrt(6+h)+ sqrt(6)

and use the idea up top that you have (a-b)(a+b) = a^2 - b^2

Answer 2

Ok let me try that

Answer 3

Ok so now I got

\[\frac{ (\sqrt{6+h}-\sqrt{6})^{2} }{ h(\sqrt{6+h}+\sqrt{6} }\]

So then the square roots cancel where I get
\[\frac { h}{ h(\sqrt{6+h}+\sqrt{6}) }\]

So I cancel the "h" on the top and bottom to get:

\[\frac{ 1 }{ \sqrt{6+h}+\sqrt{6} }\]

Answer 4

the top should be
\[ \left( \sqrt{6+h} \right)^2 - \left( \sqrt{6} \right)^2 \]

Answer 5

@Matthew071
it is sqrt(6+h)+ sqrt(6), with a plus between them for the upper term as well

Answer 6

Agreed and there fore the square roots cancel so on the top all i have remaining is H on the top, am I correct?

Answer 7

now you have
\[ \frac{1}{ \sqrt{6+h}+ \sqrt{6} } \]
take the limit as h->0

Answer 8

Ok so all I do is plug in h

Answer 9

So I get:
\[\frac{ 1 }{ 2\sqrt{6} }\]

Answer 10

yes

Answer 11

Thank you so much! Have a wonderful day!