/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 42 Show that if \(x>0\), then \(... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 42

Show that if \(x>0\), then \(e<\left(1+\frac{1}{x}\right)^{x+1}\).

Short Answer

Expert verified
To prove that if \(x > 0\), then \(e < \left(1 + \frac{1}{x}\right)^{x+1}\), we use the definition of e as the limit of \(\left(1 + \frac{1}{n}\right)^n\) as \(n\) approaches infinity. Let \(y = x+1\), then we have \(e < \left(1 + \frac{1}{y-1}\right)^{y}\), which is similar to the definition of \(e\). Since \(e\) is an upper bound for the sequence \((1 + \frac{1}{n})^n\), we conclude that \((1 + \frac{1}{y-1})^y > e\), which can be rewritten in terms of \(x\) as \(e < \left(1 + \frac{1}{x}\right)^{x+1}\).

Step by step solution

01

Use the definition of e

Recall the definition of e as the limit of the expression \((1 + \frac{1}{n})^n\) as \(n\) approaches infinity, i.e., \(e = \lim_{n \to \infty}(1 + \frac{1}{n})^n \). Note that this definition implies that for any positive integer \(n\), \( (1 + \frac{1}{n})^n \leq e \).
02

Compare the given expression with the definition of e

We want to prove that if \(x>0\), then \( e < \left(1 + \frac{1}{x}\right)^{x+1} \). Notice that if we let \(y = x+1\), then \(y - 1 = x\). Rewriting the given inequality using \(y\), we get \(e < \left(1 + \frac{1}{y-1}\right)^{y}\). We can observe now that this expression is similar to the definition of \(e\), but with \(y\) instead of \(n\).
03

Relate the inequality with the limit of the definition of e

Since \(e = \lim_{n \to \infty}(1 + \frac{1}{n})^n\), we know that \(e\) is an upper bound for the sequence \((1 + \frac{1}{n})^n\). Given that \(x>0\), we have \(y>1\) where \(y = x+1\). This means that there is a positive integer \(N\) such that \(1 < y \leq N\).
04

Compare the sequences to prove the inequality

Since \(1 < y \leq N\), we know that \((1 + \frac{1}{y-1})^y\) lies between \((1 + \frac{1}{1})^2\) and \((1 + \frac{1}{N})^N\). Since \(e\) is an upper bound for the sequence \((1 + \frac{1}{n})^n\), we can conclude that \((1 + \frac{1}{y-1})^y > e\).
05

Rewrite the inequality in terms of x

To conclude the proof, we rewrite the previous inequality in terms of \(x\) by substituting \(y = x+1\): \((1 + \frac{1}{x})^{x+1} > e\). Therefore, if \(x > 0\), then \(e < \left(1 + \frac{1}{x}\right)^{x+1}\).

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