Problem 373
Question
Evaluate the limits with either L'Hôpital's rule or previously learned methods. \(\lim _{x \rightarrow 0} \frac{(1+x)^{n}-1}{x}\)
Step-by-Step Solution
Verified Answer
The limit is \(n\).
1Step 1: Identify the Form of the Limit
First, let's substitute the value of the limit which is approaching zero. Substituting directly gives us: \ \( \frac{(1+0)^n - 1}{0} = \frac{1^n - 1}{0} = \frac{0}{0} \), which is an indeterminate form.
2Step 2: Apply L'Hôpital's Rule
Since the limit is in the indeterminate form \(\frac{0}{0}\), we can apply L'Hôpital's Rule. This rule states that \( \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \) if the limit is originally \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). Here, \(f(x) = (1+x)^n - 1\) and \(g(x) = x\).
3Step 3: Differentiate the Numerator and Denominator
Differentiate \((1+x)^n - 1\) with respect to \(x\). Using the chain rule, \(f'(x) = n(1+x)^{n-1}\). The derivative of the denominator \(g(x) = x\) is \(g'(x) = 1\).
4Step 4: Evaluate the New Limit
Now substitute the derivatives into L'Hôpital's Rule: \ \( \lim_{x \to 0} \frac{n(1+x)^{n-1}}{1} = n(1+0)^{n-1} = n \).
5Step 5: Conclude the Evaluation
By using L'Hôpital's Rule, the limit \(\lim_{x \to 0} \frac{(1+x)^n - 1}{x}\) evaluates to \(n\).
Key Concepts
Indeterminate FormsChain RuleDerivativesLimits Evaluation
Indeterminate Forms
When evaluating limits, sometimes direct substitution into the limit expression leads to results that don't immediately provide a clear answer. This is commonly known as an "indeterminate form.” Indeterminate forms occur when substituting a limit value into an expression results in 0/0 or ∞/∞. These forms are problematic because they don’t inherently inform us of the behavior of the function as it approaches the limit point.
In our example, \(\lim _{x \rightarrow 0} \frac{(1+x)^n-1}{x}\), when we substitute \(x = 0\), we get \(\frac{0}{0}\), an indeterminate form. Indeterminate forms tell us that we need more advanced techniques, like l'Hôpital's Rule, to accurately evaluate the limit.
In our example, \(\lim _{x \rightarrow 0} \frac{(1+x)^n-1}{x}\), when we substitute \(x = 0\), we get \(\frac{0}{0}\), an indeterminate form. Indeterminate forms tell us that we need more advanced techniques, like l'Hôpital's Rule, to accurately evaluate the limit.
Chain Rule
The chain rule is a fundamental technique in calculus, specifically useful in differentiation. It is employed when differentiating composite functions. A composite function is formed when one function is nested inside another. The chain rule states that if you have a function \(y = f(g(x))\), then its derivative \(y'\) is \(f'(g(x)) \cdot g'(x)\).
In the context of the given problem, we are tasked with differentiating \((1+x)^n - 1\). Here, the function \((1+x)^n\) can be considered as a composite function, where \(g(x) = 1+x\) and \(f(g(x)) = g(x)^n\). When using the chain rule, we find that the derivative of \((1+x)^n\) with respect to \(x\) is \(n(1+x)^{n-1}\).
In the context of the given problem, we are tasked with differentiating \((1+x)^n - 1\). Here, the function \((1+x)^n\) can be considered as a composite function, where \(g(x) = 1+x\) and \(f(g(x)) = g(x)^n\). When using the chain rule, we find that the derivative of \((1+x)^n\) with respect to \(x\) is \(n(1+x)^{n-1}\).
- Apply the chain rule when faced with nested functions.
- Recognize the outer and inner functions and differentiate them accordingly.
Derivatives
Derivatives represent the rate at which a function changes as its input changes. In simple terms, it measures how a function’s output changes with respect to changes in its input. The process of finding a derivative is known as differentiation.
Derivatives are central to the problem-solving process here. After identifying that our limit expression is in an indeterminate form, l'Hôpital's Rule involves taking the derivatives of the numerator and the denominator. For the function \((1+x)^n - 1\), as we've seen, the derivative is \(n(1+x)^{n-1}\), and for \(x\), it is simply \(1\). These derivatives allow us to re-evaluate the expression and smoothly compute the limit.
Derivatives are central to the problem-solving process here. After identifying that our limit expression is in an indeterminate form, l'Hôpital's Rule involves taking the derivatives of the numerator and the denominator. For the function \((1+x)^n - 1\), as we've seen, the derivative is \(n(1+x)^{n-1}\), and for \(x\), it is simply \(1\). These derivatives allow us to re-evaluate the expression and smoothly compute the limit.
Limits Evaluation
The process of limits evaluation involves determining the behavior of a function as it approaches a specific point. This concept is foundational in calculus and helps in understanding and finding function behaviors near certain points, including points where the function does not behave nicely.
In the problem, after differentiating both the numerator and the denominator, we re-evaluate the limit using these derivatives: \( \lim_{x \to 0} \frac{n(1+x)^{n-1}}{1} \). By a simple substitution of \(x = 0\) into the differentiated function, we get \(n\).
In the problem, after differentiating both the numerator and the denominator, we re-evaluate the limit using these derivatives: \( \lim_{x \to 0} \frac{n(1+x)^{n-1}}{1} \). By a simple substitution of \(x = 0\) into the differentiated function, we get \(n\).
- Recall that limits tell us how a function behaves as it approaches a certain value.
- Apply proper techniques to evaluate limits, especially when faced with indeterminate forms, often using l'Hôpital's Rule.
Other exercises in this chapter
Problem 372
For the following exercises, evaluate the limits with either L'Hôpital's rule or previously learned methods. $$ \lim _{x \rightarrow 1} \frac{x-1}{\sin x} $$
View solution Problem 373
For the following exercises, evaluate the limits with either L'Hôpital's rule or previously learned methods. $$ \lim _{x \rightarrow 0} \frac{(1+x)^{n}-1}{x} $$
View solution Problem 374
For the following exercises, evaluate the limits with either L'Hôpital's rule or previously learned methods. $$ \lim _{x \rightarrow 0} \frac{(1+x)^{n}-1-n x}{x
View solution Problem 374
Evaluate the limits with either L'Hôpital's rule or previously learned methods. \(\lim _{x \rightarrow 0} \frac{(1+x)^{n}-1-n x}{x^{2}}\)
View solution