Here's another example again if we look at it in it's

original form 3 n squared plus five divided by i n squared plus

2 i n minus 1, it's hard to see what the limit is.

Because both numerator and denominator seem to go off to infinity.

They're not bounded, they're not

convergent sequences, convergent sequences are unbounded.

But if I pull the n squared out of

numerator and denominator, I need to do that correctly.

I need to pull that out of each term, so the

numerator becomes 3 plus 5 over n squared and

the denominator becomes i plus 2 i divided by n.

There's only one n here, so we need to pull

the second n out right here, minus 1 over n squared.

Now I know 1 over n squared converges to 0, so 5 over n squared converges to

0, so the whole numerator converges to 3. In the denominator, I have i

over n squared going to 0, I have 2 i over

n going to 0 and so the denominator goes to i.

So the whole limit is 3 over i which is

not how we write complex numbers, we know by now.

Imagine the only number is in the denominator, so this is equal

to minus 3i so the limit of the sequence is minus 3i.

Here's another example. n squared

over n plus 1.

Again, both numerator and denominator go to infinity, so we try to pull out an n.

If I pull out one n, I have another n left in the numerator.

The denominator is 1 plus 1 over n.

So the denominator behaves really well. It converges to 1.

So the denominator can be treated as almost equal to 1, a margin of n.

However, the numerator gets bigger and bigger and

bigger and bigger.

And if I divide a really big number by a number that's almost 1.

The really big number doesn't change a whole lot.

And so this sequence is not bounded, it just gets bigger and bigger.

The numerator is simply stronger than the denominator.

The sequence is not bounded, and so it does not converge.

By r fact, convergence sequence are bounded.

Since this one is not bounded it cannot converge and,

finally, let's look at this example 3n plus

5 divided by in squared plus 2yn minus 1.

Again, let's pull up n squared out of both numerator and denominator,

the numerator becomes then, 3 over n plus 5 over n squared.

The denominator becomes i, plus 2 i over n minus one over n squared.

In the numerator, 3 over n goes to zero,

5 over n squared goes to 0. In the denominator, one over n squared

goes to zero, 2 i over n goes to 0 I, ios constant, equal to i.

So denominator goes to i, numerator to 0, o divided by i is 0.

So this last sequence converts this as 0, thus n goes to infinity.