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

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

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

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

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

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

something cancels

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

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

use direct substitution now!

thanks, appreciate your help both of you

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