Question

Math

Posted 7 months ago

super great

`$\lim _{x \rightarrow 0} \frac{e^{x}-e^{\sin x}}{x-\sin x}$`

Sign in to unlock the answer

Answer from Sia

Posted 7 months ago

Solution by Steps

step 1

To find the limit of the given function as $x$ approaches 0, we can use L'Hôpital's Rule, which states that if the limit of functions $f(x)$ and $g(x)$ as $x$ approaches a point $c$ results in an indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then the limit of $\frac{f(x)}{g(x)}$ as $x$ approaches $c$ is the same as the limit of $\frac{f'(x)}{g'(x)}$ as $x$ approaches $c$, provided that the latter limit exists

step 2

The function $\frac{e^x - e^{\sin(x)}}{x - \sin(x)}$ is in the indeterminate form $\frac{0}{0}$ as $x$ approaches 0. Applying L'Hôpital's Rule, we differentiate the numerator and the denominator with respect to $x$

step 3

The derivative of the numerator $e^x - e^{\sin(x)}$ with respect to $x$ is $e^x - e^{\sin(x)}\cos(x)$, and the derivative of the denominator $x - \sin(x)$ with respect to $x$ is $1 - \cos(x)$

step 4

Taking the limit of the derivatives as $x$ approaches 0, we get $\lim_{x \to 0} \frac{e^x - e^{\sin(x)}\cos(x)}{1 - \cos(x)} = \frac{1 - 1}{0} = 0$. However, this is not the final answer, as we need to apply L'Hôpital's Rule again because we still have an indeterminate form $\frac{0}{0}$

step 5

Differentiating the numerator and the denominator again, we get $\lim_{x \to 0} \frac{e^x - (e^{\sin(x)}\cos(x))'}{(1 - \cos(x))'}$. The derivative of $e^{\sin(x)}\cos(x)$ is $e^{\sin(x)}(-\sin(x)\cos(x) + \cos^2(x))$, and the derivative of $1 - \cos(x)$ is $\sin(x)$

step 6

Taking the limit of the new derivatives as $x$ approaches 0, we get $\lim_{x \to 0} \frac{e^x - e^{\sin(x)}(-\sin(x)\cos(x) + \cos^2(x))}{\sin(x)}$. As $x$ approaches 0, $e^x$ and $e^{\sin(x)}$ both approach 1, $\sin(x)$ approaches 0, and $\cos(x)$ approaches 1

step 7

Simplifying the expression, we get $\lim_{x \to 0} \frac{1 - (1)(0 + 1)}{0} = \lim_{x \to 0} \frac{1 - 1}{0} = 0$. However, this is still an indeterminate form, so we apply L'Hôpital's Rule one more time

step 8

Differentiating the numerator and the denominator one more time, we get $\lim_{x \to 0} \frac{e^x - (e^{\sin(x)}(-\sin(x)\cos(x) + \cos^2(x)))'}{\sin(x)'}$. The derivative of $\sin(x)$ is $\cos(x)$, and the derivative of the numerator simplifies to $e^x$ as $x$ approaches 0

step 9

Taking the limit of the final derivatives as $x$ approaches 0, we get $\lim_{x \to 0} \frac{e^x}{\cos(x)} = \frac{1}{1} = 1$. This is the final result of the limit

Answer

$\lim_{x \to 0} \frac{e^x - e^{\sin(x)}}{x - \sin(x)} = 1$

Key Concept

L'Hôpital's Rule and Taylor Series Expansion

Explanation

The limit was evaluated using L'Hôpital's Rule, which involves taking derivatives of the numerator and denominator until the indeterminate form is resolved. Additionally, the Taylor series expansion of $e^x$ and $e^{\sin(x)}$ around $x = 0$ confirms that the leading terms cancel out, leaving the constant term 1 as $x$ approaches 0.

Not the question you are looking for? Ask here!

Enter question by text

Enter question by image

Unlock Smarter Learning with AskSia Super!

Join Super, our all-in-one AI solution that can greatly improve your learning efficiency.

30% higher accuracy than GPT-4o

Entire learning journey support

The most student-friendly features

Study Other Question