Question

Limits problem

Answer 1

YASSSSSS!

Answer 2

multiply top and bottom by the conjugate
(to use the idea (a-b)(a+b) = a^2 - b^2

Answer 3

in other words, multiply top and bottom by \( \sqrt{8h+1} + 1 \)

Answer 4

wrong pic

Answer 5

the one i just posted is the right one

Answer 6

nothing tricky there. the bottom approaches some number
the top approaches 0
0/number is 0

Answer 7

yeah but we have to get an answer even if its 0.

Answer 8

Hold on

Answer 9

1) "Plug in" doesn't mean anything.
2) You may NOT substitute until AFTER you know it's continuous. Have you proven that?

Answer 10

I'm out to lunch on that one.

Answer 11

this is another multiply by the conjugate.
i.e. multiply top and bottom by \( \sqrt{x+22} + 19 \)

Answer 12

I did not notice that the bottom number approaches 19-19
so top and bottom are both approaching 0, and 0/0 is not defined.

However, using the conjugate trick, we can find the limit

Answer 13

\[\lim_{x \rightarrow 339}\frac{ x-339 }{ \sqrt{x+22} -19}\times \frac{ \sqrt{x+22}+19 }{ \sqrt{x+22}+19 }\]
\[=\lim_{x \rightarrow 339}\frac{ \left( x-339 \right)\left( \sqrt{x+22}+19 \right) }{ \left( x+22-361 \right) }\]
=?

Answer 14

can you finish what surji posted? simplify the bottom

Answer 15

i think i can

Answer 16

would it be 19?

Answer 17

what is x+22-361 ?

Answer 18

Leave it in terms of x

Answer 19

-339

Answer 20

x-339

Answer 21

yeah and it cancels out with the top

Answer 22

notice you now have an (x-339) in the top and bottom

as x approaches 339 (but *never* reaches it!) we can cancel (because it's not zero/zero)

Answer 23

once that annoyance is gone, we find the limit of
sqr(x+22) + 19

Answer 24

yeah and then you substitute

Answer 25

yes

Answer 26

38.

Answer 27

thanks!