Question

Limit of x-3/((root of x+6)-3) as x approaches 3?

Answer 1

try multiply the conjugate of the denominator on top and bottom

Answer 2

would the conjugate be \[\sqrt{x+6}+3\] ?

Answer 3

yep!

Answer 4

thank you!

Answer 5

so you already got the limit?

Answer 6

\[\lim_{x \rightarrow 3}\frac{x-3}{\sqrt{x+6}-3} \\ \lim_{x \rightarrow 3}\frac{(x-3)(\sqrt{x+6}+3)}{(\sqrt{x+6}-3)( \sqrt{x+6}+3)}\]
and you should see:
\[(\sqrt{x+6}-3) (\sqrt{x+6}+3)=(x+6)-9=x-3\]
so a common factor from top and bottom will "cancel"

Answer 7

And then the answer would be 6?

Answer 8

sqrt(3+6)+3
sqrt(9)+3
3+3
yes 6

Answer 9

Thank you!

Answer 10

np

Answer 11

\[\lim_{x \rightarrow 3}\dfrac{x-3}{\sqrt{x+6}-3}\]when you plug in x=3, you get 0/0, thus you can differentiate on top and bottom.
\[\lim_{x \rightarrow 3}{\Large \frac{1-0}{\frac{1}{2\sqrt{x+6}}-0}}\]
\[\lim_{x \rightarrow 3}{\Large \frac{1}{\dfrac{1}{2\sqrt{x+6}}}}\]
\[\lim_{x \rightarrow 3}2{\sqrt{x+6}}\]\[2{\sqrt{3+6}}=6\]