Question

Question #2e813

Answer 1

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

Explanation:

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.