How do you find the limit of # (sqrt(x^2+10x+1)-x)# as x approaches #oo#?

1 Answer
George C.
Sep 3, 2016

#lim_(x->oo) (sqrt(x^2+10x+1)-x) = 5#

Explanation:

Let #c# be a constant.

Then we find:

#lim_(t->oo) (sqrt(t^2+c)-t)=lim_(t->oo) (sqrt(t^2+c)-t)#

#color(white)(lim_(t->oo) (sqrt(t^2+c)-t))=lim_(t->oo) ((sqrt(t^2+c)-t)(sqrt(t^2+c)+t))/(sqrt(t^2+c)+t)#

#color(white)(lim_(t->oo) (sqrt(t^2+c)-t))=lim_(t->oo) ((t^2+c)-t^2)/(sqrt(t^2+c)+t)#

#color(white)(lim_(t->oo) (sqrt(t^2+c)-t))=lim_(t->oo) c/(sqrt(t^2+c)+t)#

#color(white)(lim_(t->oo) (sqrt(t^2+c)-t))=0#

#color(white)()#
Note that #x^2+10x+1 = (x+5)^2-24#

Let #t = x+5# and #c = -24#

Then:

#lim_(x->oo) (sqrt(x^2+10x+1)-x) = lim_(t->oo) (sqrt(t^2+c)-t+5)#

#color(white)(lim_(x->oo) (sqrt(x^2+10x+1)-x)) = lim_(t->oo) (sqrt(t^2+c)-t) + 5#

#color(white)(lim_(x->oo) (sqrt(x^2+10x+1)-x)) = 0 + 5#

#color(white)(lim_(x->oo) (sqrt(x^2+10x+1)-x)) = 5#

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