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Chapter 9: Problem 87

Proof Prove that \(\lim _{n \rightarrow \infty} \frac{x^{n}}{n !}=0\) for any real \(x\)

Short Answer

Expert verified
By breaking down the sequence and applying the squeeze theorem, it can be proved that \(\lim _{n \rightarrow \infty} \frac{x^{n}}{n !}=0\) for any real \(x\).

Step by step solution

01

Break down the sequence

First, let's break down this sequence as the ratio of two consecutive terms. To do this, observe the sequence as follows: \(\frac{x^{n+1}}{(n+1)!} \div \frac{x^n}{n!}\). By simplifying this division, you get \(\frac{x}{n+1}\)
02

Apply the squeeze theorem

Now, considering that \(x\) is any real number and \(n\) is progressing towards infinity, as \(n\) becomes infinitely large, the term \(\frac{x}{n+1}\) becomes closer and closer to 0. Ultimately, the product of a sequence of terms, each less than 1, must converge to 0. This can be shown by applying the squeeze theorem.
03

Conclusion

Since all elements of the sequence will be less than or equal to \(\frac{x}{n+1}\) which tends towards 0 as \(n\) approaches infinity, by the squeeze theorem, the sequence \(\frac{x^{n}}{n !}\) must also converge to 0 as \(n\) approaches infinity, thus proving that \(\lim _{n \rightarrow \infty} \frac{x^{n}}{n !}=0\) for any real \(x\)

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