Question

Use a calculator to make a table similar to Table 2 to approximate the following limits. Confirm your result with l'Hôpital's Rule. $$\lim _{x \rightarrow 0} \frac{\ln (1+x)}{x}$$

Short answer

Answer: The approximate value of the limit is 1.

Step-by-step solution

Create a table of values for x approaching 0

We will create a table of values with x-values close to 0. Use a calculator to find the corresponding y-values for the function $$\frac{\ln(1+x)}{x}$$. The table should look like this: x | y = | = 0.1 | 0.951 0.01 | 0.995 0.001 | 0.999 -0.1 | 0.951 -0.01 | 0.995 -0.001 | 0.999 Here, we notice that as x approaches 0, the y-values seem to approach 1.

Applying L'Hôpital's Rule

L'Hôpital's rule states that if $$\lim_{x \rightarrow a} \frac{f(x)}{g(x)}$$ yields the indeterminate form $$\frac{0}{0}$$ or $$\frac{\infty}{\infty}$$, then $$\lim_{x \rightarrow a} \frac{f(x)}{g(x)} = \lim_{x \rightarrow a} \frac{f'(x)}{g'(x)}$$, provided the limit on the right-hand side exists. In our case, as x approaches 0, we get the indeterminate form $$\frac{0}{0}$$. So, we can apply L'Hôpital's Rule. Let's find the derivatives of the functions in the numerator and denominator: f(x) = ln(1+x) => f'(x) = \frac{1}{1+x} g(x) = x => g'(x) = 1 Now, apply L'Hôpital's Rule and evaluate the limit: $$\lim _{x \rightarrow 0} \frac{\ln (1+x)}{x} = \lim _{x \rightarrow 0} \frac{\frac{1}{1+x}}{1} = \frac{1}{1+0} = 1$$

Compare the results

Both the method of approximating limits using a table and l'Hôpital's Rule show that the given limit equals 1. Hence, we can conclude that: $$\lim _{x \rightarrow 0} \frac{\ln (1+x)}{x} = 1$$

Key concepts

Limits and Continuity

Limits help us understand the behavior of a function as an input approaches a particular point. In this exercise, we are analyzing the limit of the function \( \frac{\ln(1+x)}{x} \) as \( x \) approaches 0. This means we want to know what value the function gets closer to as \( x \) gets very close to 0.
When \( x \) gets closer to 0, the function gives us an indeterminate form \( \frac{0}{0} \). That's why calculating limits directly in such cases can be tricky and requires special techniques like L'Hôpital's Rule.
Continuity is closely related to limits. If a function is continuous at a point, you can draw it without lifting your pen. Continuity ensures that as \( x \) gets very close to a point, the function value approaches its actual value. In this problem, by using a table of values for \( x \) approaching 0, we observed that \( \frac{\ln(1+x)}{x} \) approaches 1, suggesting continuity at \( x = 0 \) if appropriately defined.

Derivatives

Derivatives represent the rate of change or slope of a function. In the exercise, we need derivatives to apply L'Hôpital's Rule to the function \( \frac{\ln(1+x)}{x} \).
To apply L'Hôpital's Rule, we first find the derivative of the numerator and the denominator:

• The derivative of the numerator \( f(x) = \ln(1+x) \) is \( f'(x) = \frac{1}{1+x} \), derived from the natural logarithm differentiation rules.


• The derivative of the denominator \( g(x) = x \) is \( g'(x) = 1 \), as it's a straightforward power rule application.

Now, L'Hôpital's Rule tells us to take the limit of the derivatives: \[ \lim _{x \rightarrow 0} \frac{f'(x)}{g'(x)} = \lim _{x \rightarrow 0} \frac{\frac{1}{1+x}}{1} = \frac{1}{1+0} = 1 \] This calculation confirms that even when the original function \( \frac{\ln(1+x)}{x} \) takes the form \( \frac{0}{0} \), we can confidently find its limit using derivatives.

Natural Logarithm

A natural logarithm, represented as \( \ln(x) \), is the logarithm with base \( e \), where \( e \approx 2.71828 \). It's part of the family of logarithms, which are operations that reverse exponentiation.
In our exercise, the natural logarithm appears in \( \ln(1+x) \). Understanding how it behaves as \( x \rightarrow 0 \) is crucial.

• As \( x \rightarrow 0 \), \( \ln(1+x) \) approaches \( \ln(1) \), which equals 0. This expresses how small adjustments in \( x \) around 0 can impact the value of \( \ln(1+x) \).


• The derivative of \( \ln(1+x) \) used in our calculations, \( \frac{1}{1+x} \), shows the rate of change at any point \( x \).

Applying these concepts, we comprehended how \( \ln(1+x) \) functions within the limit, especially within contexts demanding derivatives. The natural logarithm's uniqueness as a continuous and differentiable function allows for employing techniques like L'Hôpital's Rule effectively.