To prove the limit
\[ \lim_{x \to 0} \frac{e^x - e^{-x}}{2x} = 1 \]
we will use two methods: the Taylor series expansion and the definition of the derivative. The Taylor series expansion will allow us to analyze the behavior of the function for small values of \( x \), while the definition of the derivative will provide an alternative confirmation.
Taylor Series Expansion
We use the Taylor series expansion of the exponential functions:
\[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + O(x^4) \]
\[ e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + O(x^4) \]
Subtracting the two expressions:
\[ e^x - e^{-x} = 2x + \frac{x^3}{3} + O(x^4) \]
Dividing by \(2x\):
\[ \frac{e^x - e^{-x}}{2x} = 1 + \frac{x^2}{6} + O(x^3) \]
Limit Calculation
Taking the limit as \(x\) approaches zero:
\[ \lim_{x \to 0} \frac{e^x - e^{-x}}{2x} = 1 + 0 = 1 \]
Verification Using the Definition of the Derivative
We observe that the function in the numerator is precisely the definition of the derivative of the function \( f(x) = e^x \) evaluated at \( x = 0 \):
\[ \lim_{x \to 0} \frac{e^x - e^{-x}}{2x} = \lim_{x \to 0} \frac{\sinh x}{x} \]
Since it is known that \( \lim_{x \to 0} \displaystyle \frac{\sinh x}{x} = 1 \), we obtain the result once again.