In this section, we will prove the following notable limit:
\[ \lim_{x \to 0}\frac{e^x-1}{x}=1 \]
using two distinct methods. The first is based on interpreting the limit as the definition of the derivative, while the second leverages the Taylor series expansion of the exponential function.
Interpretation as a Derivative
Consider the function \( f(x) = e^x \). The difference quotient
\[ \frac{e^x-1}{x} = \frac{e^x - e^0}{x - 0} \]
represents the difference quotient that, as \( x \) approaches 0, defines the derivative of \( f(x) \) at 0:
\[ \lim_{x \to 0}\frac{e^x-1}{x}=f'(0). \]
Since the derivative of the exponential function is itself \( e^x \), we have:
\[ f'(0)=e^0=1. \]
Therefore, we obtain:
\[ \lim_{x \to 0}\frac{e^x-1}{x}=1. \]
Taylor Series Expansion
Alternatively, we consider the Taylor series expansion of the function \( e^x \) around \( x=0 \):
\[ e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots. \]
Subtracting \( 1 \) from both sides, we obtain:
\[ e^x-1=x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots. \]
Dividing by \( x \) (for \( x \neq 0 \)):
\[ \frac{e^x-1}{x}=1+\frac{x}{2!}+\frac{x^2}{3!}+\cdots. \]
Taking the limit as \( x \to 0 \), all terms containing \( x \) vanish, yielding:
\[ \lim_{x \to 0}\left(1+\frac{x}{2!}+\frac{x^2}{3!}+\cdots\right)=1. \]
Both methods, based on the derivative definition and the Taylor series expansion, lead to the same result:
\[ \lim_{x \to 0}\frac{e^x-1}{x}=1. \]
Exercises
Exercise 1. Compute the limit:
\[ \lim_{x \to 0} \frac{e^{3x}-1}{\sin(2x)}. \]
Solution. We can rewrite the expression as:
\[ \frac{e^{3x}-1}{\sin(2x)} = \frac{e^{3x}-1}{3x} \cdot \frac{3x}{\sin(2x)}. \]
Using the notable limit for the exponential function:
\[ \lim_{x \to 0}\frac{e^{3x}-1}{3x} = 1 \]
and knowing that:
\[ \lim_{x \to 0}\frac{\sin(2x)}{2x} = 1 \quad\Rightarrow\quad \lim_{x \to 0}\frac{2x}{\sin(2x)} = 1, \]
we obtain:
\[ \lim_{x \to 0}\frac{3x}{\sin(2x)} = \lim_{x \to 0}\frac{3x}{2x} \cdot \frac{2x}{\sin(2x)} = \frac{3}{2}\cdot 1 = \frac{3}{2}. \]
Therefore:
\[ \lim_{x \to 0} \frac{e^{3x}-1}{\sin(2x)} = 1 \cdot \frac{3}{2} = \frac{3}{2}. \]
Exercise 2. Compute the limit:
\[ \lim_{x \to 0} \frac{\tan(3x)}{e^{2x}-1}. \]
Solution. We can rewrite the expression as:
\[ \frac{\tan(3x)}{e^{2x}-1} = \frac{\tan(3x)}{3x} \cdot \frac{3x}{2x} \cdot \frac{2x}{e^{2x}-1}. \]
Using the following notable limits:
\[ \lim_{x \to 0}\frac{\tan(3x)}{3x} = 1 \]
and
\[ \lim_{x \to 0}\frac{e^{2x}-1}{2x} = 1 \quad\Rightarrow\quad \lim_{x \to 0}\frac{2x}{e^{2x}-1} = 1, \]
and simplifying \( \frac{3x}{2x} \) to \( \frac{3}{2} \), we get:
\[ \lim_{x \to 0} \frac{\tan(3x)}{e^{2x}-1} = 1 \cdot \frac{3}{2} \cdot 1 = \frac{3}{2}. \]