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Mathematics
OpenStudy (anonymous):

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

myininaya (myininaya):

$\frac{1}{v}\sqrt{v+1}-\frac{1}{v}?$

OpenStudy (anonymous):

$1/v \sqrt{v+1} - 1/v$

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

myininaya (myininaya):

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

myininaya (myininaya):

something cancels

OpenStudy (anonymous):

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

myininaya (myininaya):

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

myininaya (myininaya):

use direct substitution now!

OpenStudy (anonymous):

thanks, appreciate your help both of you