Problem 41 Show that if $x>0,$ then \(... [FREE SOLUTION]

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Precalculus A Prelude to Calculus

Sheldon Axler

$Math Studyset 91影视 Explanations$ Math

1 Edition

Chapter 4: Problem 41

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

Short Answer

Expert verified

To prove that if $x > 0$, then $\left(1+\frac{1}{x}\right)^{x} 0$, we know that $f(x)$ is increasing on the interval $(0, \infty)$. By evaluating the limit of $f(x)$ as $x$ approaches infinity, we show that $f(x) < e$ when $x > 0$.

Step by step solution

Define the function and find its derivative

Let's define a function $f(x)=(1+\frac{1}{x})^{x}$. To analyze its properties, we need to find its derivative. We can do so by first taking the natural logarithm of both sides and then finding the derivative with respect to $x$. Let $y = f(x) = (1+\frac{1}{x})^{x}$. Take the natural logarithm of both sides: $\ln(y) = \ln\left((1+\frac{1}{x})^{x}\right)$ Now apply the property of logarithms: $\ln(a^b) = b\ln(a)$: $\ln(y) = x\ln(1+\frac{1}{x})$ Next, we will find the derivative of both sides with respect to $x$: $\frac{1}{y}\frac{dy}{dx} = \frac{\ln(1+\frac{1}{x}) - \frac{1}{1+x}}{x}$

Show the derivative is positive

To show the derivative is positive, let's use the fact that the natural logarithm function is increasing on the interval $(0, \infty)$: $\ln(1+\frac{1}{1+x}) > \ln(1+\frac{1}{x}) - \frac{1}{1+x}$ Now, we can recognize that $\ln(1+\frac{1}{x}) - \frac{1}{1+x}$ is the derivative of $\ln(y)$ with respect to $x$. Since $\ln(1+\frac{1}{1+x}) > \ln(1+\frac{1}{x}) - \frac{1}{1+x}$: $\frac{1}{y}\frac{dy}{dx} > 0$ Thus, $\frac{dy}{dx} > 0$ for $x > 0$. Since the derivative is positive for all $x > 0$, the function $f(x) = (1+\frac{1}{x})^{x}$ is increasing on the interval $(0, \infty)$.

Use a limit to show $f(x) < e$

As $x$ approaches infinity, we have: $\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^{x} = e$ Now, because $f(x)$ is increasing on the interval $(0, \infty)$, it follows that if $x > 0$, then $f(x) < f(\infty) = e$. Therefore, if $x > 0$, then $\left(1+\frac{1}{x}\right)^{x} < e$. This proves the given inequality.

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91影视

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

Short Answer

Step by step solution

Define the function and find its derivative

Show the derivative is positive

Use a limit to show \(f(x) < e\)

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