Problem 36 - Calculus - AP Edition

Question

Find the limits. $$\lim _{x \rightarrow 0}\left(\frac{1}{x}-\frac{1}{e^{x}-1}\right)$$

Step-by-Step Solution

Answer

The limit is infinity.

Step 1: Identifying the Expression

The limit we want to evaluate is \( \lim_{x \rightarrow 0}\left(\frac{1}{x} - \frac{1}{e^x - 1}\right) \). This involves two terms that are both singular at \( x = 0 \). This means we need to find a common form to simplify it.

Step 2: Finding a Common Denominator

Rewriting the expression with a common denominator: \[\frac{1}{x} - \frac{1}{e^x - 1} = \frac{e^x - 1 - x}{x(e^x - 1)}\].The aim is to simplify this expression to evaluate the limit.

Step 3: Using Series Expansion for Simplification

Near \( x = 0 \), we can use the Taylor series for \( e^x \):\[e^x \approx 1 + x + \frac{x^2}{2} + \cdots\]This gives us \( e^x - 1 \approx x + \frac{x^2}{2} + \cdots \). Substitute this approximation into the numerator \[e^x - 1 - x \approx \left(x + \frac{x^2}{2} + \cdots \right) - x = \frac{x^2}{2} + \cdots\].

Step 4: Simplifying the Expression and Evaluating the Limit

Substitute the approximations back:\[\lim_{x \rightarrow 0}\frac{e^x - 1 - x}{x(e^x - 1)} \approx \lim_{x \rightarrow 0}\frac{\frac{x^2}{2}}{x(x + \frac{x^2}{2})} = \lim_{x \rightarrow 0}\frac{x}{2x(x + \frac{x^2}{2})}\].This simplifies to:\[\lim_{x \rightarrow 0}\frac{1}{2(x + \frac{x^2}{2})}\approx \lim_{x \rightarrow 0}\frac{1}{2x} \approx \lim_{x \rightarrow 0} \frac{1}{2x} \rightarrow \infty\].

Step 5: Conclusion

When x approaches zero, the simplified expression, \( \frac{1}{2x} \), approaches infinity. Therefore, the original limit tends to infinity.

Key Concepts

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool in calculus for finding the limits of indeterminate forms. When you're faced with a limit that results in an indeterminate form like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), L'Hôpital's Rule can be applied to simplify the process.
This rule states that if you have a limit of the form \( \lim_{{x \to c}} \frac{f(x)}{g(x)} \), where both \( f(x) \) and \( g(x) \) approach zero or both approach infinity as \( x \to c \), then you can differentiate the numerator and the denominator separately:

• Compute the derivatives: \( f'(x) \) and \( g'(x) \).
• Then find \( \lim_{{x \to c}} \frac{f'(x)}{g'(x)} \).

If the new limit exists, it is equal to the original limit.
This method can be repeated if the resulting limit is still indeterminate. However, it's crucial that the functions are differentiable near \( c \) and that \( g'(x) eq 0 \) when \( x \to c \). L'Hôpital's Rule simplifies complex limit problems into manageable derivatives, making it an essential tool in calculus.

Taylor Series Expansion

The Taylor series expansion is a method used to approximate complex functions using polynomials. It is especially useful when we need to analyze functions around a specific point.
The Taylor series for a function \( f(x) \) centered at a point \( a \) is given by:
\[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots\]
This enables you to approximate \( f(x) \) by polynomials when \( x \) is close to \( a \).
For calculating limits, Taylor series are useful as they allow us to express complex expressions in terms of simpler polynomials.
For example, the exponential function can be expanded as:
\[e^x \approx 1 + x + \frac{x^2}{2} + \frac{x^3}{3!} + \cdots\]
Near \( x=0 \), this expansion becomes particularly handy in limit calculations, as it shows the behavior of \( e^x - 1 \) as \( x \to 0 \). The polynomial form reveals insights into the function's behavior that aren't immediately apparent.
This simplified view helps you perform algebraic manipulations to find limits easily.

Indeterminate Forms

Indeterminate forms arise in calculus when evaluating certain limits, where substitution leads to an ambiguous result. These forms make it unclear how the limit behaves at that point.
Common indeterminate forms include:

• \( \frac{0}{0} \)
• \( \frac{\infty}{\infty} \)
• \( 0 \cdot \infty \)
• \( \infty - \infty \)
• \( 0^0 \)
• \( \infty^0 \)
• \( 1^\infty \)

If a limit results in any of these forms, it suggests that more analysis is required using algebraic manipulation, L'Hôpital's Rule, or other techniques such as Taylor series.
For instance, the exercise involves a limit that initially presents a \( \frac{0}{0} \) form. This signals the need for further simplification or the application of L'Hôpital's Rule for resolution.
Understanding indeterminate forms is crucial for correctly solving limits.
They highlight the need to think beyond direct substitution and ensure that the correct mathematical methods are applied to uncover limit behavior.