How do you find the limit of #sqrt(9x^2 +x)-(3x)# as x approaches infinity?

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Answer 1 · 2016-09-06T16:59:04

# = 1/6#

#lim_(x to oo) sqrt(9x^2 +x)-3x#

#lim_(x to oo) 3x sqrt(1 +1/(9x))-3x#

#lim_(x to oo) 3x( (1 +1/(9x))^(1/2)-1)#

by Binomial Expansion

#lim_(x to oo) 3x( 1 + 1/2 1/(9x) + O(1/x^2) -1)#

#lim_(x to oo) 3x( 1/(18x) + O(1/x^2) )#

#lim_(x to oo) 1/6 + O(1/x) = 1/6#

Answer 2 · 2016-09-06T17:18:10

#1/6#

#((sqrt(9x^2+x)-3x))/1 * ((sqrt(9x^2+x)-3x))/((sqrt(9x^2+x)-3x)) = (9x^2+x-9x^2)/((sqrt(9x^2+x)-3x))#

# = x/(sqrt(9x^2+x)-3x)#

# = x/(sqrt(x^2)sqrt(9+1/x) - 3x)# #" "# for #x != 0#

# = x/(x(sqrt(9+1/x) - 3))#

# = 1/(sqrt(9+1/x) - 3)#

Now, as #x rarr oo#, #1/x rarr 0#, so the limit is

#1/(sqrt9 + 3) = 1/6#