Question

Can someone please explain: limit as v approaches 0 of: (1/v(sqrt v+1)-(1/v)

Answer 1

\[\frac{1}{v}\sqrt{v+1}-\frac{1}{v}?\]

Answer 2

\[1/v \sqrt{v+1} - 1/v\]

Answer 3

Get a common denominator:
\[\lim_{v \rightarrow 0}\frac{1}{v \sqrt{v+1}}-\frac{1}{v}=\lim_{v \rightarrow 0}\frac{1-\sqrt{v+1}}{v \sqrt{v+1}}\]
You see that gives you 0/0 so use l'hospital's rule:
\[\lim_{v \rightarrow 0}\frac{-\frac{1}{2}(v+1)^{\frac{-1}{2}}}{\sqrt{v+1}+v \frac{1}{2}(v+1)^{\frac{-1}{2}}}\]
(taking the derivative of top and bottom)
Then you can plug in zero and see the limit is -1/2

Answer 4

what if i havent yet learned l'hospitals rule, how can i approach this algebraically

Answer 5

\[\frac{1}{v*\sqrt{v+1}}-\frac{1}{v}=\frac{1-\sqrt{v+1}}{v*\sqrt{v+1}}=\frac{1-\sqrt{v+1}}{v*\sqrt{v+1}}*\frac{1+\sqrt{v+1}}{1+\sqrt{v+1}}\]
\[=\frac{1-(v+1)}{v*\sqrt{v+1}*(1+\sqrt{v+1}}=\frac{-v}{v*\sqrt{v+1}*(1+\sqrt{v+1})}\]

Answer 6

something cancels

Answer 7

That is the alternative I was going to type up^^ But I like L'hospital's lol

Answer 8

\[\frac{-1}{\sqrt{v+1}*(1+\sqrt{v+1})}\]

Answer 9

use direct substitution now!

Answer 10

thanks, appreciate your help both of you