Chapter 6: Problem 46
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
Expert verified
Answer: The approximate value of the limit is 1.
Step by step solution
01
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.
02
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:
$$\lim _{x \rightarrow 0} \frac{\ln (1+x)}{x} = \lim _{x \rightarrow 0} \frac{\frac{1}{1+x}}{1} = \frac{1}{1+0} = 1$$
03
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:
$$\lim _{x \rightarrow 0} \frac{\ln (1+x)}{x} = 1$$
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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:
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.
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.
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 \( x \).