We want to prove the following limit:
\[ \lim_{x \to 0} (1 + x)^{\frac{1}{x}} = e \]
We will prove this limit using two distinct methods: the natural logarithm and its derivative, and the Taylor series expansion.
Natural Logarithm and Derivative
Let us set:
\[ y = (1 + x)^{\frac{1}{x}} \]
Applying the natural logarithm:
\[ \ln y = \frac{1}{x} \ln(1 + x) \]
We compute the limit of the right-hand side:
\[ \lim_{x \to 0} \frac{\ln(1 + x)}{x} \]
Using the fundamental limit:
\[ \lim_{x \to 0} \frac{\ln(1 + x)}{x} = 1 \]
We obtain:
\[ \lim_{x \to 0} \ln y = 1 \]
Which implies:
\[ \lim_{x \to 0} y = e \]
Taylor Series Expansion
We use the Taylor series expansion for \( \ln(1 + x) \) around \( x = 0 \):
\[ \ln(1 + x) = x - \frac{x^2}{2} + O(x^3) \]
Thus, we have:
\[ \frac{\ln(1 + x)}{x} = 1 - \frac{x}{2} + O(x^2) \]
Taking the limit as \( x \to 0 \), we obtain:
\[ \lim_{x \to 0} \frac{\ln(1 + x)}{x} = 1 \]
Therefore:
\[ \lim_{x \to 0} \ln y = 1 \]
Which implies:
\[ \lim_{x \to 0} y = e \]
Thus, we have proven the limit using two distinct approaches.
Exercises
Using the remarkable limit, compute the following limits:
Exercise 1. Compute:
\[ \lim_{x \to 0} \left( 1 + 2x \right)^{\frac{1}{x}} \]
Solution. Let \( y = (1 + 2x)^{\frac{1}{x}} \), and take the logarithm:
\[ \ln y = \frac{1}{x} \ln(1 + 2x) \]
By the fundamental limit:
\[ \lim_{x \to 0} \frac{\ln(1 + 2x)}{x} = 2 \]
Thus:
\[ \lim_{x \to 0} \ln y = 2 \Rightarrow \lim_{x \to 0} y = e^2 \]
Exercise 2. Compute:
\[ \lim_{x \to 0} \left( 1 + x^2 \right)^{\frac{1}{x^2}} \]
Solution. Let \( y = (1 + x^2)^{\frac{1}{x^2}} \) and take the logarithm:
\[ \ln y = \frac{1}{x^2} \ln(1 + x^2) \]
Using the Taylor series expansion:
\[ \ln(1 + x^2) = x^2 - \frac{x^4}{2} + O(x^6) \]
We obtain:
\[ \frac{\ln(1 + x^2)}{x^2} = 1 - \frac{x^2}{2} + O(x^4) \]
Taking the limit:
\[ \lim_{x \to 0} \ln y = 1 \Rightarrow \lim_{x \to 0} y = e \]