Problem 45 Combine to show that \(\left(1... [FREE SOLUTION]

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Precalculus A Prelude to Calculus

Sheldon Axler

$Math Studyset 91影视 Explanations$ Math

2 Edition

Chapter 3: Problem 45

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

Short Answer

Expert verified

For $x>0$, we first show that $\left(1+\frac{1}{x}\right)^{x}1$. Together, these inequalities give us $\left(1+\frac{1}{x}\right)^{x}0$.

Step by step solution

Show that $\left(1+\frac{1}{x}\right)^{x}0$.

Since $x>0$, we can rewrite the inequality as: \[\left(1+\frac{1}{x}\right)^{x}\infty}\left(1+\frac{1}{x}\right)^{x} \leq e.\] Now, we need to analyze the limit expression on the left side. According to the limit definition of the exponential function, we have: \[\lim_{x->\infty}\left(1+\frac{1}{x}\right)^{x} = e.\] Thus, for all $x>0$, we have: \[\left(1+\frac{1}{x}\right)^{x}0,\) then $e<\left(1+\frac{1}{x}\right)^{x+1}$.

Show $e0$.

Let's assume $x$ is a positive real number. We need to show that \[e<\left(1+\frac{1}{x}\right)^{x+1}.\] Now, using the property $a^{m+n} = a^m \cdot a^n$, we can rewrite the expression on the right side as: \[\left(1+\frac{1}{x}\right)^{x}\left(1+\frac{1}{x}\right).\] Since we have already shown in step 1 that $\left(1+\frac{1}{x}\right)^{x}0$, and because $1+\frac{1}{x}>1$ for $x>0$, the product of these two factors will be greater than $e$. Therefore, \[e<\left(1+\frac{1}{x}\right)^{x}\left(1+\frac{1}{x}\right)=\left(1+\frac{1}{x}\right)^{x+1}\] for $x>0$. Now putting both parts together, we get \[\left(1+\frac{1}{x}\right)^{x}0\).

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91影视

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

Short Answer

Step by step solution

Show that \(\left(1+\frac{1}{x}\right)^{x}0\).

Show \(e0\).

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