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Answer 1 · 2017-12-17T01:20:17

$lim_"x ->oo" sqrt(x+1)/(e^sqrtx)=0$

This is of the form $lim_"x ->oo" sqrt(x+1)/(e^sqrtx)=oo/oo$

Applying L'Hospitals rule:

$lim_"x ->oo" sqrt(x+1)/(e^sqrtx)=lim_"x ->oo" (1/(2sqrt(x+1)))/((e^sqrtx)(1/(2sqrt(x))))$

$lim_"x ->oo" (2sqrt(x))/((e^sqrtx)(2sqrt(x+1)))=lim_"x ->oo" (cancel2sqrt(x))/((e^sqrtx)(cancel2sqrt(x+1))$

$lim_"x ->oo" ((sqrt(x))/((e^sqrtx)(sqrt(x+1))))*((sqrt(1/x))/(sqrt(1/x)))$

$lim_"x ->oo" (sqrt(1))/((e^sqrtx)(sqrt(1+1/x)))=lim_"x ->oo" (1)/((e^sqrtx)(sqrt(1+1/x)))=1/(e^oo*sqrt(1+0))=1/oo=0$

God bless....I hope the explanation is useful.