Compare the rates of growth of f(x) = the square root of x and g(x) = Ln(x) as x approaches infinity.

Question

Compare the rates of growth of f(x) =  and g(x) = Ln(x) as x approaches infinity.

 f(x) grows faster than g(x) as x goes to infinity.

 g(x) grows faster than f(x) as x goes to infinity. 

f(x) and g(x) grow at the same rate as x goes to infinity. 

The rate of growth cannot be determined.

Thank you so much in advance!

Answer 1

Ws're comparing the rate of change of the two functions, which is dy/dx, the first derivative.

The derivative of f(x) is 1/2 x^(-1/2). This is the rate of change of f(x)

The derivative of g (x) is 1/x . This is the rate of change of g(x).

An easy way to compare these two is to substitute a very large number for x, say 1,000,000

1/2 (1,000,000)^(-1/2)= 1/2000= .0005

1/1,000,000 = .000001

So f(x) is changing faster, although both curves are very flat!

If we want to see what are the values of x where f' is greater than g' , set up and solve this inequality:

1/2 x^(-1/2) > 1/x

There are many ways to solve this! Here's one. We're looking at x>0, so we can muliply each side by 2x, then square both sides, without changing the inequality. It's really hard to type out all of the algebra, so I'll jump to the end:

x>4 For all x>4 , f(x) increases faster!

Answer 2

limit f(x) / g(x) is indeterminate. By L'Hopital's rule, lim f(x) / g(x) = lim f'(x) / g'(x)

= lim (1/2 x^(-1/2) / (1/x) = lim (sqrt x)/ 2 which is infinite.

So, f(x) grows faster.