Calculus 2: Sequence Limits Question to the power n?

raaznar
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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


(\frac{3^{n}-4^{n}}{3n^{2}+4^{n}+7})

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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raaznar said:

Homework Statement


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

Homework Equations


(\frac{3^{n}-4^{n}}{3n^{2}+4^{n}+7})

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 → ∞?
 
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 4^n will 'dominate' here.

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

Now, what does each of those fractions go to?
 
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.
 
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 4^n will 'dominate' here.

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

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!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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