Find the limit of this function?

Question

Limit as x-->infinity of sqrt(2x^2+1)/3x-5
This is an example in my book and they've divided numerator and denominator by x and then it became Limit as x-->infinity of sqrt(2+1/x^2)/3-5/x

I don't understand why the numerator is sqrt(2+1/x^2). Please explain. That's all I need to know. Thanks.

thanks puggy

Answer 1

lim sqrt(2x^2 + 1)/(3x - 5)
x -> infinity

Factor out x^2 inside the square root.

lim sqrt( x^2 (2 + 1/x^2) ) / (3x - 5)
x -> infinity

x^2 gets pulled out as |x|, from the square root.

lim |x| sqrt( 2 + (1/x^2)) / (3x - 5 )
x -> infinity

|x| = x as x approaches infinity.

lim x sqrt( 2 + (1/x^2)) / (3x - 5 )
x -> infinity

Divide top and bottom by x.

lim sqrt(2 + (1/x^2)) / ( 3 - 5/x )
x -> infinity

Which can now be evaluated directly.

sqrt(2 + 0) / (3 - 0)
sqrt(2)/3

Answer 2

(1/x)√(2x^2 + 1) =
√(1/x^2)(2x^2 + 1) =
√(2x^2/x^2 + 1/x^2) =
√(2 + 1/x^2)