find the limits: 1.f(x)=sqrt(x^2+4x+3) -(x+2) when x goes to +infinity 2. f(x)=[sqrt(x+6) - 3]/ (x-3) x=3 any help is appreciated. thanx
Last edited by mr fantastic; Oct 5th 2009 at 05:50 PM. Reason: Moved from another thread
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1. multiply top and bottom by ($\displaystyle \sqrt{x^2+4x+3}+(x+2)$) then let xgo to infinity
yeah, i've tried that and f(x)=1/ sqrt(x^2+4x+3) + x+2 and it's still an indertiminate form sorry for the bad english
Originally Posted by blertta yeah, i've tried that and f(x)=1/ sqrt(x^2+4x+3) + x+2 and it's still an indertiminate form sorry for the bad english yes thats right so when x goes to infinity sqrt(x^2+4x+3) + x+2 goes to infinity so you get 1/h h goes to infinity and this limit is zero try to write the equations properly so that we know what is being divided. you can do the same thing to 2 aswell.
Last edited by mr fantastic; Oct 5th 2009 at 05:51 PM. Reason: Merged posts and edited
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