Evaluating Limits Without L'Hopital's Rule: A Comprehensive Guide

When faced with evaluating certain limits, particularly indeterminate forms like 0/0 or infin;/infin;, one common approach is L'Hopital's Rule. However, there are alternative methods that do not rely on differentiation. This article explores the evaluation of the limit (lim_{x to 0} frac{ln(1 x)}{x}); we will delve into the Taylor series expansion of (ln(1 x)) and other methods that offer a deeper understanding of limit evaluation without resorting to L'Hopital's Rule.

Using Taylor Series Expansion

One effective way to evaluate the limit without L'Hopital's Rule is to use the Taylor series expansion of (ln(1 x)) around (x 0). The Taylor series for (ln(1 x)) is given by:

$$ln(1 x) x - frac{x^2}{2} frac{x^3}{3} - frac{x^4}{4} cdots$$

for (|x| 1).

By substituting this series into the given limit, we get:

(lim_{x to 0} frac{ln(1 x)}{x} lim_{x to 0} frac{x - frac{x^2}{2} frac{x^3}{3} - frac{x^4}{4} cdots}{x})

This simplifies to:

(lim_{x to 0} left(1 - frac{x}{2} frac{x^2}{3} - frac{x^3}{4} cdots right))

As (x) approaches (0), the terms (-frac{x}{2}), (frac{x^2}{3}), etc., all tend to (0). Therefore, the limit evaluates to:

(boxed{1})

Alternative Approach: Bernoulli's Inequality

Another method to approach this problem is by using Bernoulli's inequality. Let's prove it by induction for ((-1 y 1)) and then apply the definition of the exponential function.

Given ((1 y)^n geq 1 ny), we can use this to show that:

(e^t lim_{n to infty} left(1 frac{t}{n}right)^n leq 1 t)

Now, substitute (t ln(1 x)).

(1 x e^{ln(1 x)} leq 1 ln(1 x))

Therefore, for (x 0), we have:

(frac{1}{x} leq ln(1 x))

On the other hand, for (x 0) and (u 1 x), where (u to 1) as (x to 0):

(-ln(1 x) lnleft(frac{1}{1 x}right) lnleft(1 - frac{x}{1 x}right) leq frac{x}{1 x})

This implies:

(frac{x}{x 1} leq ln(1 x) leq x)

By the Squeeze theorem, we conclude:

(lim_{x to 0} frac{ln(1 x)}{x} 1)

Limit Evaluation Using Variable Substitution and Tangent Line

A final approach is through variable substitution and the concept of the tangent line. Let's set (x 1 u), so (x u - 1) and as (x to 0), (u to 1). This transforms the given limit into:

(lim_{x to 0} frac{ln(1 x)}{x} lim_{u to 1} frac{ln u}{u-1})

Consider the function (f(x) ln x). The tangent line to (f(x)) at the point ((1, 0)) has the equation:

(y - f(1) f'(1)(x - 1))

Given that (f(1) 0) and (f'(1) 1), the equation simplifies to:

(y x - 1)

Since the values of the tangent line approach the values of the function (ln x) close to (x 1), we have:

(lim_{x to 1} ln x lim_{x to 1} (x - 1))

Dividing both sides by (x - 1), we get:

(frac{lim_{x to 1} ln x}{lim_{x to 1} (x - 1)} 1)

Thus, (lim_{x to 1} frac{ln x}{x - 1} 1). Therefore:

(boxed{1})