How do you find #lim sqrt(2x+3)-sqrtx# as #x->oo#?

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Answer 1 · 2018-01-18T16:17:16

#lim_(xrarr+oo)(sqrt(2x+3)-sqrtx)=+oo#

#lim_(xrarr+oo)(sqrt(2x+3)-sqrtx)=?#

#f(x)=sqrt(2x+3)-sqrtx=#

#((sqrt(2x+3)-sqrtx)(sqrt(2x+3)+sqrtx))/(sqrt(2x+3)+sqrtx)# #=#

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

#(2x+3-x)/(sqrt(2x+3)+sqrtx)# #=#

#(x+3)/(sqrt(2x+3)+sqrtx)#

As a result,

#lim_(xrarr+oo)(sqrt(2x+3)-sqrtx)=lim_(xrarr+oo)(x+3)/(sqrt(2x+3)+sqrtx)# #=#

#lim_(xrarr+oo)(x+3)/(sqrt(x^2(2/x+3/x^2))+sqrtx)# #=#

#lim_(xrarr+oo)(x+3)/(|x|sqrt(2/x+3/x^2)+|x|sqrt(1/x))# #=#

#x->+oo#
#x>0#

#lim_(xrarr+oo)(x+3)/(xsqrt(2/x+3/x^2)+xsqrt(1/x))# #=#

#lim_(xrarr+oo)(cancel(x)(1+3/x))/(cancel(x)(sqrt(2/x+3/x^2)+sqrt(1/x))# #=#

#lim_(xrarr+oo)(1+3/x)/(sqrt(2/x+3/x^2)+sqrt(1/x))=^((1/0^+)# #+oo#