Question

Fun Calculus Question! find numbers a and b
lim sqrt{ax+b}-2 =1
x->0 ----------
x

Answer 1

\[\lim_{x \rightarrow 0}\frac{\sqrt{ax+b}-2}{x}=1\]

Answer 2

How far did you get on this prob before you got stuck? :)

Answer 3

i just started doing it. i figure it would be a fun problem for everyone to attemp since it is in the problems plus section of my calculus book

Answer 4

Cool. Well, I got the answer in about 2 seconds using Geogebra and some sliders. To do it analytically might take a little longer. :)

Answer 5

i think i saw this somewhere before,,,,but i'm sure i never solved it...it's fun to try now :D

Answer 6

myininaya, if you wanna see the geogebra answer I can share a screen with you on my virtual whiteboard.

Answer 7

sure! lets do it

Answer 8

what did you get for a and b?

Answer 9

a=4 b=4? is that right let me check

Answer 10

im gonna join :D if you dont mind

Answer 11

Hope that was enjoyable, play with geogebra, it's awesome! :D

Answer 12

ok since the bottom goes to 0 the top would have to go to 0
\[\sqrt{ax+b}-2=0\] => x->0, b=4
\[\lim_{x \rightarrow 0}a \frac{1}{2}(ax+b)^\frac{-1}{2}=\lim_{x \rightarrow 0}\frac{a}{2\sqrt{ax+b}}=\frac{a}{2\sqrt{b}}\] but b=4
\[\frac{a}{2*2}=\frac{a}{4}=1\] => a=4

Answer 13

I like that approach myininaya

Answer 14

:)

Answer 15

somewhere kinda unseen to me, but i'm still curious how you get the second limit :)

Answer 16

you mean a?

Answer 17

a/2squr(ax+b) but how you approach to that

Answer 18

L' hospital rule?

Answer 19

yes

Answer 20

we know the limit is suppose to be 1
so thats why we set a/4=1

Answer 21

oh ok, i see it now...Great and fun :)