How do you find the limit of #sqrt(x^2+7x-6)+x# as x approaches #oo#?

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Answer 1 · 2018-05-02T00:27:29

The answer #lim_(xrarroo)sqrt(x^2+7x-6)+x=oo#

show below

#lim_(xrarroo)sqrt(x^2+7x-6)+x#

Notes that

#(oo)^2=oo#

#oo+oo=oo#

#c+oo=oo#

#-c+oo=oo#

#sqrt(oo)=oo#

#c*oo=oo#

#-c*oo=-oo#

now let solve it

#lim_(xrarroo)sqrt(x^2+7x-6)+x=sqrt[(oo)^2+7(oo)-6]+oo=oo#