 Indeterminate Form (Part 1) How to solve the limits in indeterminate form [(infinity)/(infinity), (infinity) - (infinity)]: examples and their solutions.

Example 1: ∞/∞ As n → ∞,
both of the numerator and the denominator go to ∞.

And the highest order of the numerator (3n2)
and the hightest order of the denominator (n2)
are the same: n2.

In this case,
focus on the coefficients of the highest order term.

So (given) = 3/1
= 3.

To see why this method is true,
let's see the 'official' way to solve this example.

The highest order term is n2.

So divide both of the numerator and the denominator by n2.

Recall that 1/∞ = 0

Limits of sequences

Then (given) = (3 + 0 + 0)/(1 + 0).

You can see that
only the highest order term's coefficients remain:
3 and 1.

So 3/1 = 3 is the answer.

This is why
we focused on the highest term's coefficient
on the left side solution.

Example 2: ∞/∞ First, arrange the exponets
to clearly see the bases of the exponential terms.

There are three types of terms:
3n, 4n, and n100.

As n → ∞,
all of these terms go to ∞.

But 4n is the fastest term that goes to ∞.

(Exponential functions whose base is greater than 1
goes to ∞ faster than any polynomial functions.
So n100 is the slowest term in this case.)

So focus on the fastest term's coefficients:
(given) = 2/1
= 2.

To see why this method is true,
let's see the 'official' way to solve this example.

First, arrange the exponets
to clearly see the bases of the exponential terms.

The fastest term is 4n.

So divide both of the numerator and the denominator by 4n.

Recall that 1/∞ = 0

Limits of sequences

So (given) = (0 + 2 - 0)/(1 - 0 + 0).

You can see that
only the highest order term's coefficients remain:
2 and 1.

So 2/1 = 2 is the answer.

Example 3: ∞/∞ The hightest order term in the numerator (3n)
and the highest order terms in the denominator (√4n2, √n2)
have the same order.

So, as n → ∞,
(given) = 3/(√4 + √1).

Example 4: ∞ - ∞ The denominator is in ∞ - ∞ form.

And the highest order term of √n2 + n (√n2)
and the highest order term of √n2 - n (√n2)
have the same order.

Then multiply the denominator's conjugate,
n2 + n + √n2 - n,
to both of the numerator and the denominator.

Rationalizing a denominator

Cancel the dark gray terms.

The hightest order terms in the numerator (√n2, √n2)
and the highest order term in the denominator (2n)
have the same order.

So, as n → ∞,
(given) = [7(√1 + √1)] / 2.

Example 5: ∞ - ∞ The denominator is in ∞ - ∞ form.

And the highest order term of √n2 + 6n + 12 (√n2)
and the highest order term of n (n)
have the same order.

So multiply the given expression's conjugate,
n2 + 6n + 12 + n,
to both of the numerator and the denominator.

The hightest order term in the numerator (6n)
and the highest order terms in the denominator (√n2, n)
have the same order.

So, as n → ∞,
(given) = 6/(√1 + 1).