How do you find the limit of #(sqrt(x+6)-x)/(x^3-3x^2)# as #x->-oo#?

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Answer 1 · 2018-02-06T10:51:53

#0#

#(sqrt(x+6)-x)/(x^3-3x^2)#

Divide by #x^3#:

#((sqrt(x+6))/x^3-x/x^3)/(x^3/x^3-3x^2/x^3)=((sqrt(x+6))/x^3-1/x^2)/(1-3/x)#

#lim_(x->-oo)((sqrt(x+6))/x^3-1/x^2)/(1-3/x)=(lim_(x->-oo)((sqrt(x+6))/x^3-1/x^2))/(lim_(x->-oo)(1-3/x)#

#lim_(x->-oo)((sqrt(x+6))/x^3-1/x^2)=lim_(x->-oo)(sqrt(x+6))/x^3-lim_(x->-oo)(1/x^2)#

#lim_(x->-oo)(sqrt(x+6))/x^3=0#

#lim_(x->-oo)(1/x^2)=0#

#0-0=0#

#lim_(x->-oo)(1-3/x)=lim_(x->-oo)(1)-lim_(x->-oo)(3/x)#

#lim_(x->-oo)(1)=1#

#lim_(x->-oo)(3/x)=0#

#:.#

#1-0=1#

#0/1=0#

Hence:

#lim_(x->-oo)(sqrt(x+6)-x)/(x^3-3x^2)=0#