ass $3$ : ( $4$ ) fuints) $\newline$ Une the Mean Value Theorem to establish the following inequality for $x>0$ : $\newline$ $1<\frac{e^{x}-1}{\ln (x+1)}<(x+1) e^{x}$ $\newline$ Doduce the following limit : $\newline$ $\lim _{x \rightarrow 0} \frac{e^{x}-1}{\ln (x+1)}$

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Algebra 2

Simplify radical expressions with root inside the root

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Q. ass

3

: (

4

) fuints)

\newline

Une the Mean Value Theorem to establish the following inequality for

x>0

\newline

1<\frac{e^{x}-1}{\ln (x+1)}<(x+1) e^{x}

\newline

Doduce the following limit :

\newline

\lim _{x \rightarrow 0} \frac{e^{x}-1}{\ln (x+1)}

Function Analysis: Let's consider the function $f(x) = e^x - 1$ on the interval $[0, x]$ where $x > 0$ . According to the Mean Value Theorem, there exists some $c$ in $(0, x)$ such that $f'(c) = \frac{f(x) - f(0)}{x - 0}$ .
Derivative Calculation: First, we calculate $f'(x)$ which is the derivative of $f(x) = e^x - 1$ . The derivative $f'(x)$ is $e^x$ .
Mean Value Theorem Application: Now, applying the Mean Value Theorem, we get $e^c = \frac{e^x - 1}{x}$ for some $c$ in $(0, x)$ . Since $e^x$ is an increasing function, we have $e^0 < e^c < e^x$ , which means $1 < e^c < e^x$ .
Inequality Formation: Multiplying all sides of the inequality by $x$ , we get $x < x\cdot e^c < x\cdot e^x$ . Since $e^c = \frac{e^x - 1}{x}$ , we can substitute to get $x < e^x - 1 < x\cdot e^x$ .
Function Analysis: Now, let's consider the function $g(x) = \ln(x+1)$ on the interval $[0, x]$ . We apply the Mean Value Theorem again to find that there exists some $d$ in $(0, x)$ such that $g'(d) = \frac{g(x) - g(0)}{x - 0}$ .
Derivative Calculation: We calculate $g'(x)$ , which is the derivative of $g(x) = \ln(x+1)$ . The derivative $g'(x)$ is $\frac{1}{x+1}$ .
Mean Value Theorem Application: Applying the Mean Value Theorem to $g(x)$ , we get $\frac{1}{d+1} = \frac{\ln(x+1) - \ln(1)}{x} = \frac{\ln(x+1)}{x}$ for some $d$ in $(0, x)$ . Since $\frac{1}{x+1}$ is a decreasing function, we have $\frac{1}{x+1} < \frac{1}{d+1} < 1$ .
Inequality Formation: Multiplying all sides of the inequality by $x$ , we get $\frac{x}{x+1} < \ln(x+1) < x$ . Since $\frac{1}{d+1} = \frac{\ln(x+1)}{x}$ , we can substitute to get $\frac{x}{x+1} < \ln(x+1) < x$ .
Combined Inequalities: Now, we combine the inequalities we have for $e^x - 1$ and $ext{ln}(x+1)$ . We have $x < e^x - 1 < x*e^x$ and rac{x}{x+1} < ext{ln}(x+1) < x. Dividing the first inequality by the second, we get rac{rac{x}{x+1}}{x} < rac{e^x - 1}{ ext{ln}(x+1)} < rac{x*e^x}{x}/rac{x+1}{x}.
Limit Deduction: Simplifying the inequality, we get $\frac{1}{x+1} < \frac{e^x - 1}{\ln(x+1)} < e^x$ . Since $x > 0$ , we can multiply all sides by $x+1$ to get $$1$ < \frac{e^x - $1$}{\ln(x+$1$)} < (x+$1$)e^x.
L'Hôpital's Rule Application: To deduce the limit as $x$ approaches $0$, we observe that both the numerator $e^x - 1$ and the denominator $\ln(x+1)$ approach $0$ as $x$ approaches $0$. By applying L'Hôpital's Rule, we can find the limit of the ratio as $x$ approaches $0$.
Limit Calculation: Applying L'Hôpital's Rule, we differentiate the numerator and the denominator. The derivative of the numerator $e^x - 1$ is $e^x$, and the derivative of the denominator $ ext{ln}(x+1)$ is $1/(x+1)$.
Limit Calculation: Applying L'Hôpital's Rule, we differentiate the numerator and the denominator. The derivative of the numerator $e^x - 1$ is $e^x$, and the derivative of the denominator $ ext{ln}(x+1)$ is $1/(x+1)$.Taking the limit as $x$ approaches $0$ of the derivatives, we get $ ext{lim}_{x o 0} rac{e^x}{1/(x+1)} = ext{lim}_{x o 0} e^x imes (x+1) = 1 imes (0+1) = 1$.
Limit Calculation: Applying L'Hôpital's Rule, we differentiate the numerator and the denominator. The derivative of the numerator $e^x - 1$ is $e^x$, and the derivative of the denominator $ ext{ln}(x+1)$ is $1/(x+1)$.Taking the limit as $x$ approaches $0$ of the derivatives, we get $ ext{lim}_{x \to 0} \frac{e^x}{1/(x+1)} = ext{lim}_{x \to 0} e^x * (x+1) = 1 * (0+1) = 1$.Therefore, the limit of $rac{e^x - 1}{ ext{ln}(x+1)}$ as $x$ approaches $0$ is $e^x$$0$.