How do you find the limit of #(sqrt(x^2+1))/(3x-1)# as x approaches infinity?

2 Answers
Massimiliano
Feb 22, 2015

The answer are:

for #xrarr+oo#, the limit is #1/3#,

For #xrarr-oo#, the limit is #-1/3#.

This is because:

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

#=lim_(xrarr+-oo)|x|/(3x)=(1)#

For #xrarr+oo#, #|x|=x#,

for #xrarr-oo#, #|x|=-x#.

So:

#(1)=lim_(xrarr+oo)x/(3x)=1/3#,

#(1)=lim_(xrarr-oo)(-x)/(3x)=-1/3#.

Babette
Feb 22, 2015

We can easily find the limit by dividing every term by #x# as follows
Remember when we put x into the radical we must square it
I am assuming positive infinity. In other words x greater than or equal to zero.

#(sqrt{x^2/x^2+1/x^2})/((3x)/x-1/x) #

#(sqrt{1+1/x^2})/(3-1/x) #

Now taking the limit as x approaches infinity we have

#(sqrt{1+0})/(3-0)=sqrt{1}/3=1/3 #

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