How do you find #lim sqrt(3x+1)-2sqrtx# as #x->oo#?

1 Answer
NickTheTurtle
Dec 12, 2017

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

Explanation:

The question asks to find #lim_(x->oo)(sqrt(3x+1)-2sqrt(x))#.

Multiply by #(sqrt(3x+1)+2sqrt(x))/(sqrt(3x+1)+2sqrt(x))# to get
#=lim_(x->oo)((sqrt(3x+1)-2sqrt(x))*(sqrt(3x+1)+2sqrt(x))/(sqrt(3x+1)+2sqrt(x)))#
#=lim_(x->oo)((3x+1-4x)/(sqrt(3x+1)+2sqrt(x)))#
#=lim_(x->oo)((1-x)/(sqrt(3x+1)+2sqrt(x)))#
#=lim_(x->oo)((1-x)/(sqrt(x)(sqrt(3+1/x)+2)))#
#=lim_(x->oo)((1/sqrt(x)-sqrt(x))/(sqrt(3+1/x)+2))#
#=(lim_(x->oo)(1/sqrt(x)-sqrt(x)))/(lim_(x->oo)(sqrt(3+1/x)+2))#

The numerator evaluates to #-oo# while the denominator is a finite number.

Thus, the answer is #-oo#.

This can be seen clearly with a graph:
graph{sqrt(3x+1)-2sqrt(x) [-1, 100, -5, 2]}

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