Problem 374 - Calculus Volume 1

Question

Evaluate the limits with either L'Hôpital's rule or previously learned methods. \(\lim _{x \rightarrow 0} \frac{(1+x)^{n}-1-n x}{x^{2}}\)

Step-by-Step Solution

Answer

The limit is \(\frac{n(n-1)}{2}\).

Step 1: Analyze the Indeterminate Form

First, substitute \(x = 0\) into the expression \(\frac{(1+x)^n - 1 - nx}{x^2}\). This gives \(\frac{(1+0)^n - 1 - n \cdot 0}{0^2} = \frac{0}{0}\), which is an indeterminate form. Hence, L'Hôpital's Rule or expansion methods can be used.

Step 2: Use Binomial Series Expansion

Expand \((1 + x)^n\) using the binomial series expansion: \[(1 + x)^n = 1 + nx + \frac{n(n-1)}{2}x^2 + \cdots\]Substitute this into the original limit expression:\[\frac{(1+x)^n - 1 - nx}{x^2} = \frac{1 + nx + \frac{n(n-1)}{2}x^2 + \cdots - 1 - nx}{x^2}\]Cancel out \(1\) and \(nx\):\[\frac{\frac{n(n-1)}{2}x^2 + \cdots}{x^2}\]

Step 3: Simplify the Limit Expression

The simplified expression is \[\frac{n(n-1)}{2} + \frac{\text{higher order terms}}{x^2}\]. As \(x\) approaches \(0\), the higher order terms tend to zero, simplifying the limit to \[\frac{n(n-1)}{2}\].

Step 4: Evaluate the Limit

Since the higher order terms vanish as \(x\to 0\), the limit becomes:\[\lim _{x \rightarrow 0} \frac{(1+x)^{n}-1-nx}{x^{2}} = \frac{n(n-1)}{2}\].

Key Concepts

Indeterminate Forms

When evaluating limits, you might come across expressions that seem unsolvable at first glance—these are often indeterminate forms. An indeterminate form happens when the limit leads to an ambiguous result, like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).

In the given problem, substituting \( x = 0 \) into \( \frac{(1+x)^n - 1 - nx}{x^2} \) results in \( \frac{0}{0} \). This qualifies as an indeterminate form, which means the limit isn't straightforwardly solvable by normal substitution.

• There are several types of indeterminate forms, including \( \frac{0}{0} \), \( 0 \times \infty \), and \( \infty - \infty \).
• Special techniques, like L'Hôpital's Rule or series expansions, help resolve these forms.

Recognizing an indeterminate form is your cue to apply advanced calculus methods to find the limit.

Binomial Series Expansion

Binomial series expansion is a powerful tool for converting complicated expressions into a manageable series of terms. For expressions of the form \((1+x)^n\), the binomial series helps in breaking down the powers of \(x\) into separate terms.

The binomial series expansion is written as: \[ (1 + x)^n = 1 + nx + \frac{n(n-1)}{2}x^2 + \cdots \]

• It expands the power into a sum of individual terms.
• Each term involves higher powers of \(x\), getting smaller as the power increases.

By substituting these terms back into the original expression, we simplify the entire limit evaluation process. Canceling out similar terms like \(1\) and \(nx\) reduces complexity further, essentially focusing on terms that are necessary to solve the problem.

Limit Evaluation

Once you've expanded the expression using a binomial series, the next step is limit evaluation. After substituting and simplifying the expression, we're left with: \[ \frac{\frac{n(n-1)}{2}x^2 + \cdots}{x^2} \]

• Cancellation of non-essential parts leads us to work with the expression: \( \frac{n(n-1)}{2} \).
• Higher order terms in the series expansion become negligible as \(x\) approaches zero.

The crucial part is identifying which terms become zero and which terms remain, giving us the final value of the limit as \( x \to 0 \). Ultimately, the limit evaluates to \( \frac{n(n-1)}{2} \).

Calculus Methods

In calculus, we encounter several methods tailored to solve limits involving indeterminate forms. Two such methods are L'Hôpital's Rule and series expansion.

L'Hôpital's Rule states that if you have an indeterminate form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), you can differentiate the numerator and the denominator and then take the limit again. This works when both functions are differentiable.

• L'Hôpital's Rule is especially handy when functions contain roots or logarithms.
• This approach often simplifies the form quickly.

However, for problems involving series or polynomial expressions, the binomial series expansion can often simplify the problem without differentiation, letting you directly evaluate the limit. Using the right calculus method based on the problem structure makes solving limits faster and more intuitive.