Calculus 2: Sequence Limits Question to the power n?

In summary, the conversation is about finding the limits of sequences whose nth term is given by a specific equation. The conversation includes a discussion on how to approach such problems, with the suggestion to consider the "dominate term" approach and to check for factors such as exponentials and growing denominators. The solution is -1.
  • #1
raaznar
6
0
Calculus 2: Sequence Limits Question to the power n??

Homework Statement


Find the limits (if it exists) to decide which sequences, whose nth term is given below.

Homework Equations


[itex](\frac{3^{n}-4^{n}}{3n^{2}+4^{n}+7})[/itex]

The Attempt at a Solution


I've done a few of these but as Soon as the constants were raised to the n. It pretty much stumped me. I haven't done stuff like this yet and I'm sort of self teaching.

I would appreciate some direction rather than an answer. Thanks!
 
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  • #2
raaznar said:

Homework Statement


Find the limits (if it exists) to decide which sequences, whose nth term is given below.

Homework Equations


[itex](\frac{3^{n}-4^{n}}{3n^{2}+4^{n}+7})[/itex]

The Attempt at a Solution


I've done a few of these but as Soon as the constants were raised to the n. It pretty much stumped me. I haven't done stuff like this yet and I'm sort of self teaching.

I would appreciate some direction rather than an answer. Thanks!

Notice the quantity in your numerator is always smaller than zero.

Notice your denominator is growing without bound.

So what does the whole thing tend to as n → ∞?
 
  • #3
As n grows, exponentials, numbers to the n power, will increase faster than n to a power. And, of course, a larger base will increase faster than a smaller base. That means that [tex]4^n[/tex] will 'dominate' here.

Divide each term in numerator and denominator by [itex]4^n[/itex] to get
[tex]\frac{\left(\frac{3}{4}\right)^n- 1}{3\frac{n^2}{4^n}+ 1+ \frac{7}{4^n}}[/tex]

Now, what does each of those fractions go to?
 
  • #4
Zondrina said:
Notice the quantity in your numerator is always smaller than zero.

Notice your denominator is growing without bound.

So what does the whole thing tend to as n → ∞?

0? I got 0 to start of with but the answer is -1. If that's what you are implying. I have no idea how to mathematically arrive at the answer -1.
 
  • #5
HallsofIvy said:
As n grows, exponentials, numbers to the n power, will increase faster than n to a power. And, of course, a larger base will increase faster than a smaller base. That means that [tex]4^n[/tex] will 'dominate' here.

Divide each term in numerator and denominator by [itex]4^n[/itex] to get
[tex]\frac{\left(\frac{3}{4}\right)^n- 1}{3\frac{n^2}{4^n}+ 1+ \frac{7}{4^n}}[/tex]

Now, what does each of those fractions go to?

Negative 1. Got it! I've never done the 'Dominate Term' Approach. Should this be the first approach to consider when doing limits of sequences?

What's the first things I should consider when I approach questions like this in the future? (Whats the checklist to look out for)

Thanks man by the way!
 

1. What are sequence limits in Calculus 2?

Sequence limits in Calculus 2 refer to the behavior of a sequence as it approaches a particular value or "limit". This is an important concept in understanding the convergence or divergence of a sequence.

2. How do you find the limit of a sequence in Calculus 2?

To find the limit of a sequence in Calculus 2, you can use the following steps: 1. Determine the pattern of the sequence and write out the first few terms. 2. Take the limit of each term as n approaches infinity. 3. If the limit of the terms is a constant number, then that is the limit of the sequence. 4. If the limit does not exist or approaches infinity, then the sequence is said to diverge.

3. What is the difference between a convergent and divergent sequence in Calculus 2?

A convergent sequence in Calculus 2 is one that has a finite limit as n approaches infinity. This means that the terms of the sequence get closer and closer to a particular value as n increases. A divergent sequence, on the other hand, does not have a finite limit and either approaches infinity or does not have a defined limit at all.

4. How do you prove that a sequence is convergent in Calculus 2?

To prove that a sequence is convergent in Calculus 2, you can use one of the following methods: 1. Use the definition of a limit to show that the terms of the sequence get closer and closer to a particular value. 2. Use the squeeze theorem, which states that if a sequence is bounded by two other sequences with the same limit, then the original sequence also has that limit. 3. Use a known convergence test, such as the ratio test or comparison test, to show that the sequence converges.

5. Can a divergent sequence in Calculus 2 have a finite limit?

No, a divergent sequence in Calculus 2 cannot have a finite limit. The definition of a divergent sequence is one that does not have a finite limit, either approaching infinity or not having a defined limit at all.

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