I need some help with Limits (calculus).?

Question

Find limit as n goes to infinity of f(n)/g(n)

a. f(n) = 2^(root n), g(n) = n^((log n)^2)

b. f(n) = n/log n, g(n) = log n log log n

c. f(n) = ((16^log n)/(n^2)+1), g(n) = (n^5 + 10 (root n))/ 2n^3


Also, if possible, could you provide short explanations as to how you got the answer (to refresh my memory)

Thank you so very very much.

Answer 1

These problems are basically about knowing the relative growth rate of functions -- the main things you have to remember are that all polynomials dominate all exponentials, and that given two polynomials, the one with the greatest exponent on the leading term will dominate the other one (unless they have the same greatest exponent, in which case the limit is determined by the quotient of the coefficients of the greatest terms only). Remembering these general principles, these problems become easy:

a: [n→∞]lim (2^√n)/(n^((ln n)²))

First, rewrite this in terms of e:

[n→∞]lim (e^(ln 2 √n))/(e^((ln n)³))

Use the laws of exponents:

[n→∞]lim e^(ln 2 √n - (ln n)³)

Of course, √n grows much faster than any power of ln n, so the exponent diverges to infinity, and therefore so does the entire expression.

b: [n→∞]lim (n/ln n)/(ln n ln ln n) = [n→∞]lim n/((ln n)² ln ln n)

The denominator grows no faster than (ln n)³, and again any polynomial will grow faster than any power of the logarithm.

c: [n→∞]lim (16^(ln n)/(n^2)+1) / ((n^5+10√n)/(2n³))

First simplify:

[n→∞]lim 2 (n 16^(ln n)+n³) / (n^5+10√n)

Now, write the exponent in terms of e:

[n→∞]lim 2 (n e^(ln 16 ln n)+n³) / (n^5+10√n)

Now write it as a polynomial:

[n→∞]lim 2 (n n^(ln 16)+n³) / (n^5+10√n)
[n→∞]lim 2 (n^(ln 16 + 1)+n³) / (n^5+10√n)

Now, ln 16 + 1 is less than 5, so the highest power in this rational function is 5, and it dominates all powers in the numerator. As such, the limit is zero.