The Limit of ln(1 x) - x as x Approaches 0: Understanding the Indeterminate Form and Resolving It

Understanding the limit of functions as they approach certain values can be a challenging yet fascinating journey in calculus. Today, we delve into the specific case of the limit: limx→0 ln(1 x) - x. This limit presents an indeterminate form, namely 0/0, and we will explore several methods to resolve it.

Introduction to Indeterminate Forms

An indeterminate form is a mathematical expression that cannot be evaluated directly because it represents a situation where standard algebraic methods fail to provide a clear answer. Common indeterminate forms include 0/0, ∞/∞, 0·∞, ∞ - ∞, 00, 1∞, and ∞0. In this case, as we approach 0, both the numerator (ln(1 x) - x) and the denominator (1) tend to 0, leading to the indeterminate form 0/0. This is where advanced techniques come into play.

Using L'H?pital's Rule

Step-by-Step Application of L'H?pital's Rule

L'H?pital's rule is a powerful method to resolve indeterminate forms of the type 0/0 or ∞/∞. The rule states that if the limit of the ratio of two functions is of the indeterminate form 0/0 or ∞/∞, then the limit of the ratio of their derivatives is the same, provided the limit of the derivatives exists.

Applying L'H?pital's rule to our problem:

limx→0 (ln(1   x) - x) / x
 limx→0 (d/dx[ln(1   x) - x]) / (d/dx[x])
 limx→0 (1/(1   x) - 1) / 1
 limx→0 (1/(1   x) - 1)

Simplifying the expression further:

limx→0 (1 - (1   x)) / (1   x)  limx→0 -x / (1   x)  0

However, this still results in the indeterminate form 0/0. We need to differentiate again:

limx→0 (1 - (1   x)) / (1   x)  limx→0 -1 / (1   x)  -1 / 1  -1

This still doesn't resolve the form. Let's differentiate the numerator and the denominator again:

limx→0 -1 / (1   x)  limx→0 -1 / (1   x)  -1 / 1  0 / 1  0

The correct limit is:

limx→0 -1 / (1   x)  -1 / 1  -1

Since the problem is resolved with a limit of 0, let's simplify the expression correctly. The correct form after simplification is:

limx→0 (ln(1   x) - x) / x  0

Therefore, the final result using L'H?pital's rule is:

limx→0 (ln(1 x) - x) / x 0

Using Taylor Series Expansion

Another method to resolve this indeterminate form is through Taylor series expansion. The Taylor series expansion of ln(1 x) around x 0 is given by:

ln(1   x)  x - x2/2   x3/3 - ...

Substituting this expansion into the limit, we get:

limx→0 (ln(1   x) - x) / x  limx→0 (x - x2/2 - x) / x
 limx→0 (-x2/2) / x
 limx→0 -x/2
 0

Thus, the limit of ln(1 x) - x as x approaches 0 is 0.

Conclusion and Further Exploration

Understanding limits and indeterminate forms is crucial for advanced calculus. The methods discussed here, including L'H?pital's rule and Taylor series expansion, are powerful tools for resolving such complex problems. The final result of the limit is 0, and the graph of the function confirms this conclusion.