To find the limit [tex]\(\lim _{x \rightarrow 0} \frac{e^x-1}{x}\)[/tex], follow these steps:
1. Understand the problem: We need to find the limit as [tex]\( x \)[/tex] approaches 0 of the function [tex]\( \frac{e^x - 1}{x} \)[/tex].
2. Apply L'Hôpital's Rule: The indeterminate form [tex]\(\frac{0}{0}\)[/tex] suggests that we can use L'Hôpital's Rule. L'Hôpital's Rule states that for limits of the form [tex]\(\frac{0}{0}\)[/tex] or [tex]\(\frac{\infty}{\infty}\)[/tex], the limit of [tex]\(\frac{f(x)}{g(x)}\)[/tex] as [tex]\( x \to a \)[/tex] can be found by taking the derivatives of the numerator and the denominator:
[tex]\[
\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}
\][/tex]
3. Differentiate the numerator and denominator:
- The derivative of the numerator [tex]\( e^x - 1 \)[/tex] is [tex]\( e^x \)[/tex].
- The derivative of the denominator [tex]\( x \)[/tex] is 1.
4. Apply the derivatives to L'Hôpital's Rule:
[tex]\[
\lim_{x \to 0} \frac{e^x - 1}{x} = \lim_{x \to 0} \frac{e^x}{1}
\][/tex]
5. Evaluate the limit:
- As [tex]\( x \)[/tex] approaches 0, [tex]\( e^x \)[/tex] approaches [tex]\( e^0 = 1 \)[/tex].
[tex]\[
\lim_{x \to 0} e^x = 1
\][/tex]
Therefore, the limit exists and is equal to 1. Thus, we have:
[tex]\[
\lim _{x \rightarrow 0} \frac{e^x-1}{x} = 1
\][/tex]
