Wednesday, 7 January 2015

calculus - Evaluate the limxtoinfty(x+sqrtx2+2x)




Evaluate : limx (x+x2+2x)



I've tried some basic algebraic manipulation to get it into a form where I can apply L'Hopital's Rule, but it's still going to be indeterminate form.



This is what I've done so far



limx (x+x2+2x)=limx (x+x2+2x)(xx2+2xxx2+2x)=limx (x2(x2+2x)xx2+2x)=limx (2xxx2+2x)



And that's as far as I've gotten. I've tried applying L'Hopitals Rule, but it still results in an indeterminate form.



Plugging it into WolframAlpha shows that the correct answer is 1



Any suggestions on what to do next?


Answer




limx(2xxx2+2x)=limx(2xxx2(1+2x))=limx2xx|x|1+2x==limx2xx+x1+2x=limx2xx(1+1+2x)=1


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