Question

Limit Help!
Rationalize the numerator to find the exact value of the limit analytically:

f(x)= the square root of (2x+1) minus the square root of 3
all of that over (x-1)

Answer 1

\[\frac{\sqrt{2x+1}-3}{x-1}\]?

Answer 2

root (2x+1) - root 3

Answer 3

do you know what it means when it says "rationalize the numerator"?
it means multiply top and bottom by \(\sqrt{2x+1}+\sqrt3\)
leave the denominator in factored form

Answer 4

the numerator will be
\[2x+1-3=2x-2=2(x-1)\]

Answer 5

the denominator will be
\[(x-1)(\sqrt{2x+1}+\sqrt3)\]

Answer 6

i got that much, but I didn't know how to solve the denominator

Answer 7

leave it be

Answer 8

you get
\[\frac{2(x-1)}{(x-1)(\sqrt{2x+1}+3)}\]

Answer 9

then cancel and get
\[\frac{2}{\sqrt{2x+1}+\sqrt3}\]

Answer 10

then replace \(x\) by \(1\) and you are done

Answer 11

why do you replace x with 1?

Answer 12

didn't you want the limit as \(x\to 1\)?

Answer 13

it didn't give me what x was approaching - it just said "Rationalize the numerator to find the exact value of the limit analytically"

Answer 14

the answer in the back of the book says the limit is root 3 divided by 3

Answer 15

up until this point
\[\frac{2(x-1)}{(x-1)(\sqrt{2x+1}+3)}\] you cannot replace \(x\) by \(1\) because you will get \(\frac{0}{0}\)
but once you get to \[\frac{2}{\sqrt{2x+1}+\sqrt3}\] you can
you get just what it has in the back of the book

Answer 16

it doesn't give me what to replace x with is what I'm saying

Answer 17

oh wait. I'm sorry - I didn't see the top of the page. x approaches 1+

Answer 18

@satellite73