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Chapter 9: Problem 10

Find the limit of the following sequences or determine that the limit does not exist. $$\left\\{\frac{n^{12}}{3 n^{12}+4}\right\\}$$

Short Answer

Expert verified
Answer: The limit of the sequence as \(n\) approaches infinity is \(\frac{1}{3}\).

Step by step solution

01

Consider the given sequence

The sequence we are given is \({\displaystyle \left\{\frac{n^{12}}{3 n^{12}+4}\right\}}\).
02

Manipulate the sequence by dividing both the numerator and the denominator by \(n^{12}\).

To simplify the sequence, we are going to divide the numerator and the denominator by the highest power of \(n\), which is \(n^{12}\). Doing this, we get: $$\left\{\frac{n^{12}}{3 n^{12}+4}\right\}\cdot\frac{1/n^{12}}{1/n^{12}} = \left\{\frac{1}{3+\frac{4}{n^{12}}}\right\}$$
03

Find the limit as \(n\) approaches infinity

Now we want to find the limit as n approaches infinity: $$\lim_{n\to\infty}\frac{1}{3+\frac{4}{n^{12}}}$$ As n approaches infinity, the term \(\frac{4}{n^{12}}\) approaches 0, since the denominator grows much faster than the numerator. Hence, we get: $$\lim_{n\to\infty}\frac{1}{3+\frac{4}{n^{12}}} = \frac{1}{3+0} = \frac{1}{3}$$ So the limit of the sequence as \(n\) approaches infinity is \(\frac{1}{3}\).

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