In this demonstration, we will calculate the notable limit:
\[ \lim_{x \to 0} \frac{(1+x)^{\alpha} - 1}{x} = \alpha. \]
To prove this limit, we will use the Taylor series expansion around \( x = 0 \) and the definition of the derivative to calculate the limit as the derivative of the function \( (1+x)^{\alpha} \) at \( x = 0 \).
Taylor Series Expansion
Let’s begin with the first approach, using the Taylor series of the function \( (1+x)^{\alpha} \) around the point \( x = 0 \).
The Taylor series of \( (1+x)^{\alpha} \) is given by:
\[ (1+x)^{\alpha} = 1 + \alpha x + \frac{\alpha(\alpha-1)}{2!} x^2 + \frac{\alpha(\alpha-1)(\alpha-2)}{3!} x^3 + \dots \]
Now we substitute this expansion into the limit expression:
\[ \frac{(1+x)^{\alpha} - 1}{x} = \frac{1 + \alpha x + \frac{\alpha(\alpha-1)}{2!} x^2 + \dots - 1}{x} \]
Simplifying:
\[ \frac{\alpha x + \frac{\alpha(\alpha-1)}{2!} x^2 + \dots}{x} = \alpha + \frac{\alpha(\alpha-1)}{2!} x + \dots \]
As \( x \to 0 \), all terms with \( x \) tend to zero, leaving only:
\[ \lim_{x \to 0} \frac{(1+x)^{\alpha} - 1}{x} = \alpha. \]
Definition of Derivative
Another way to prove the limit is by using the definition of the derivative. Let’s consider the function \( f(x) = (1+x)^{\alpha} \). The derivative of \( f(x) \) at \( x = 0 \) is defined as:
\[ f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h}. \]
In our case, \( f(x) = (1+x)^{\alpha} \), so:
\[ f'(0) = \lim_{h \to 0} \frac{(1+h)^{\alpha} - (1+0)^{\alpha}}{h} = \lim_{h \to 0} \frac{(1+h)^{\alpha} - 1}{h}. \]
This is exactly the form of the limit we need to calculate! The derivative of the function \( f(x) = (1+x)^{\alpha} \) is:
\[ f'(x) = \alpha (1+x)^{\alpha-1}. \]
So, evaluating at \( x = 0 \):
\[ f'(0) = \alpha (1+0)^{\alpha-1} = \alpha. \]
Therefore, we have:
\[ \lim_{x \to 0} \frac{(1+x)^{\alpha} - 1}{x} = \alpha. \]
We have proven the notable limit:
\[ \lim_{x \to 0} \frac{(1+x)^{\alpha} - 1}{x} = \alpha, \] using both the Taylor series expansion and the definition of the derivative.