How do you find the limit of #((4x^2+3x)^(1/2))-2x# as x approaches infinity?

1 Answer
Jim H
Sep 9, 2016

Please see the explanation section below.

Explanation:

#sqrt(4x^2+3x)-2x = ((sqrt(4x^2+3x)-2x)) /1 * ((sqrt(4x^2+3x)+2x)) /((sqrt(4x^2+3x)+2x)) #

# = (4x^2+3x-4x^2)/(sqrt(4x^2+3x)+2x)#

# = (3x)/(sqrt(x^2)sqrt(4+3/x)+2x)# #" "# for #x != 0#

For #x > 0#, we have #sqrt(x^2) = x#, so

# = (3x)/(xsqrt(4+3/x)+2x)# #" "# for #x > 0#

# = (3x)/(x(sqrt(4+3/x)+2))# #" "# for #x > 0#

# = (3)/(sqrt(4+3/x)+2)# #" "# for #x > 0#.

Now as #xrarroo#, we get

#(3)/(sqrt(4+0)+2) = 3/4#

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