In this page, we will prove the following limit:
\[ \lim_{x \to 0} \frac{\sinh x}{x} = 1 \]
We will explore three different methods to obtain this result. In particular, we will use the definition of the hyperbolic sine and a fundamental exponential limit, apply L'Hôpital's Theorem, and finally use the Taylor series expansion.
- Proof using the definition and fundamental limits
- Proof using L'Hôpital's theorem
- Proof using Taylor series expansion
Proof using the definition and fundamental limits
The hyperbolic sine is defined as:
\[ \sinh x = \frac{e^x - e^{-x}}{2} \]
Dividing by \( x \):
\[ \frac{\sinh x}{x} = \frac{e^x - e^{-x}}{2x} \]
We can rewrite the numerator as:
\[ e^x - e^{-x} = e^x - \frac{1}{e^x} \]
Multiplying numerator and denominator by \( e^x \):
\[ \frac{e^{2x} - 1}{2x e^x} \]
We use the fundamental limit:
\[ \lim_{y \to 0} \frac{e^y - 1}{y} = 1, \quad \text{with } y = 2x \]
We obtain:
\[ \lim_{x \to 0} \frac{e^{2x} - 1}{2x e^x} = \frac{1}{e^0} = 1 \]
Proof using L'Hôpital's theorem
We observe that the limit presents an indeterminate form \( \frac{0}{0} \), so we can apply L'Hôpital's theorem.
We differentiate the numerator and denominator:
\[ \frac{d}{dx} (e^x - e^{-x}) = e^x + e^{-x}, \quad \frac{d}{dx} (2x) = 2 \]
Thus:
\[ \lim_{x \to 0} \frac{e^x - e^{-x}}{2x} = \lim_{x \to 0} \frac{e^x + e^{-x}}{2} = 1 \]
Proof using Taylor series expansion
We expand the exponential functions into Taylor series around \( x = 0 \):
\[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + O(x^4) \]
\[ e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + O(x^4) \]
Computing the difference:
\[ e^x - e^{-x} = 2x + \frac{2x^3}{3!} + O(x^5) \]
Dividing everything by 2:
\[ \sinh x = x + \frac{x^3}{3!} + O(x^5) \]
Dividing by \( x \):
\[ \frac{\sinh x}{x} = 1 + \frac{x^2}{3!} + O(x^4) \]
Taking the limit as \( x \to 0 \), all terms containing \( x \) vanish, and we obtain:
\[ \lim_{x \to 0} \frac{\sinh x}{x} = 1 \]