#A= lim_"x->oo" sqrt((x^2+4)/(x+4))# #B=lim_"x->oo" sqrt((x^2+5)/x^3)# Find #A and B# ?

Answer 1

Ezra Adams

See below

#lim_ (x -> oo) sqrt((x^2+4)/(x+4)#

#= lim_ (x -> oo) sqrt((1+4/x^2)/(1/x+4/x^2)#

#= sqrt((1+lim_ (x -> oo)4/x^2)/(lim_ (x -> oo)1/x+ lim_ (x -> oo)4/x^2)#

#=sqrt((1+ 0)/( 0+ 0))#

#= oo#

#lim_ (x-> oo) sqrt((x^2+5)/x^3)#

#= lim_ (x-> oo) sqrt(1/x + 5/x^3)#

#= sqrt(lim_ (x-> oo) 1/x + lim_ (x-> oo) 5/x^3)#

#= sqrt 0 #

#= 0#

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Answer 2

Charles Anctil

#oo# and #0#

#lim_ (x -> oo) sqrt((x^2+4)/(x+4)) = sqrt(lim_ (x -> oo)(x^2/x)(1+4/x^2)/(1+4/x)) = oo#

#lim_ (x-> oo) sqrt((x^2+5)/x^3) = sqrt(lim_ (x-> oo)(x^2)/(x^3)(1+5/x^2)) = 0#

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Answer 3

Charles Anctil

#oo, 0#

#A= lim_"x->oo" sqrt((x^2+4)/(x+4))#

#= lim_"x->oo" sqrt((x+4/x)/(1+4/x))#

#=lim_"x->oo" sqrt(x/1) = oo#

#B=lim_"x->oo" sqrt((x^2+5)/x^3)#

#= lim_"x->oo" sqrt((1+5/x^2)/x)#

#=lim_"x->oo" sqrt(1/x) = 0#

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Answer 4

Ezra Adams

A = 1 B = 0

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

#A= lim_"x-&gt;oo" sqrt((x^2+4)/(x+4))# #B=lim_"x-&gt;oo" sqrt((x^2+5)/x^3)# Find #A and B# ?

#A= lim_"x->oo" sqrt((x^2+4)/(x+4))# #B=lim_"x->oo" sqrt((x^2+5)/x^3)# Find #A and B# ?