Question

Help: Limitation Problems

Answer 1

With due respect Welcome to Open Study ...... Sir/Ma'am @viemer

Answer 2

\[\lim_{u \rightarrow 2}[ \sqrt{4u+1}-3] / u-2\]

Answer 3

thank you ^^

Answer 4

do you get \(\frac{0}{0}\) when you replace \(u\) by \(2\) ?

Answer 5

YES... but the answer on the book is different

Answer 6

if so, rationalize the numerator by multiplying top and bottom by
\(\sqrt{2u+1}+3\)

Answer 7

the \(u-2\) will cancel when you do it

Answer 8

lol of course the answer in the book is different, \(\frac{0}{0}\) is not a number
as such it is not the answer to anything
i was just asking, because if you get \(\frac{0}{0}\) then there is more work to do, whereas if you get say \(\frac{3}{4}\) then that would be the answer

Answer 9

is it clear how to finish it now, or you need more explanation?

Answer 10

shouldn't it be it's conjugate sq (4u-1) +3?

Answer 11

sq (4u+1)+3

Answer 12

second one

Answer 13

you get \(4u+1-9=4u+8=4(u-2)\) then cancel the \(u-2\) top and bottom

Answer 14

so will it be \[4/ \sqrt{4u+1} +3 ?\]

Answer 15

I am pretty sure you have to use "L'Hôpital's rule".

@viemer: have you covered L'Hôpital's rule?

Answer 16

I ddon't know what's that but I my answer here is 2/3

Answer 17

i bet this problem precedes l'hopital by 4 weeks at least

Answer 18

@viemer yes, the answer is what you wrote

Answer 19

then replace \(u\) by \(2\) and you will get \(\frac{2}{3}\)

Answer 20

thank you!!! You're a great help. I just had a hard time because I kept on simplifying the denominator

Answer 21

well actually you will get \(\frac{4}{6}\) but that is the same thing

Answer 22

Alright, if you have not covered, then @satellite is correct.

Answer 23

oh right, i should have mentioned not to do that!!

Answer 24

leave the denominator in factored form, so you can cancel
don't multiply out, you will get a big mess