To find the limit of the function [tex]\(\lim _{x \rightarrow 0} \frac{e^x-e^{-x}}{2 x}\)[/tex], we need to analyze the behavior of the expression as [tex]\(x\)[/tex] approaches 0. Here's a step-by-step approach to understand this limit:
1. Function Definition:
[tex]\[
\lim _{x \rightarrow 0} \frac{e^x - e^{-x}}{2x}
\][/tex]
2. Check Substitution:
Directly substituting [tex]\(x = 0\)[/tex] into the function, we get:
[tex]\[
\frac{e^0 - e^0}{2 \cdot 0} = \frac{1 - 1}{0} = \frac{0}{0}
\][/tex]
This form is indeterminate (0/0), meaning we need to use other techniques to evaluate the limit.
3. Use L’Hôpital's Rule:
When faced with an indeterminate form of the type [tex]\(\frac{0}{0}\)[/tex], we can use L’Hôpital's Rule, which states that:
[tex]\[
\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}, \text{ provided the limit on the right side exists}
\][/tex]
For our function, let [tex]\(f(x) = e^x - e^{-x}\)[/tex] and [tex]\(g(x) = 2x\)[/tex].
4. Differentiate the Numerator and Denominator:
[tex]\[
f'(x) = (e^x - e^{-x})' = e^x - (-e^{-x}) = e^x + e^{-x}
\][/tex]
[tex]\[
g'(x) = (2x)' = 2
\][/tex]
5. Apply L’Hôpital’s Rule:
[tex]\[
\lim_{x \to 0} \frac{e^x - e^{-x}}{2x} = \lim_{x \to 0} \frac{e^x + e^{-x}}{2}
\][/tex]
6. Evaluate the New Limit:
Now plug [tex]\(x = 0\)[/tex] into the simplified limit:
[tex]\[
\frac{e^0 + e^0}{2} = \frac{1 + 1}{2} = \frac{2}{2} = 1
\][/tex]
Thus, the limit of the given function as [tex]\(x\)[/tex] approaches 0 is:
[tex]\[
\boxed{1}
\][/tex]
