# Someone help. Sequence and continuous functions.

1. Aug 29, 2009

### Charles007

I am confused with sequence and continuous functions.

I am confued with their limit. how do they know the min and max before they attempt the question. and is that the only solution to the question? I mean. Everytimes if I see kind question like this, is that only way to do it?...

Many thx.

Squence

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Continuous Functions.

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2. Aug 29, 2009

### mathman

The short answer is that they don't know the details in advance. Using the first problem as an example, you see that for large n, the numerator is essentially n2, while the denominator is essentially n3, so the fraction is roughly 1/n. The rest is just filling in details to make it rigorous.

3. Aug 30, 2009

### Charles007

I still don't understand how to get it.

4. Aug 30, 2009

### HallsofIvy

Do you understand that the example, as given, is not an example on how to find the limit but how to prove the limit is correct after you have found it? That is, for the purposes of the example, how the limit, 0, is found, is irrelevant.

The first sequence is
$$\left\{\frac{n^2+ 2n}{n^3- 5}\right\}$$
and, as mathman said, for very large n, the highest power in both numerator and denominator will be far larger than the rest so, for large n, the fraction will be close to $n^2/n^3= 1/n$ which goes to 0 as n gets larger and larger.

So we guess that the limit of the given sequence is 0. Then we proceed to show, as in the example, that 0 is, in fact, correct.

Last edited by a moderator: Aug 30, 2009
5. Aug 30, 2009

### slider142

There is no hard and fast rule. Limits are a way of rigorising the art of approximation. Your job is to simplify an expression enough so that you see the general behavior as it approaches some point, and then guess or approximate what happens at that point by comparing it to functions you already know. You can also use the rigor of making sure that there is some number bounding the expression.
This is partially why you studied the behavior of all those different types of functions in precalculus.