Problem 35 - Thomas Calculus

Question

Use series to evaluate the limits. \begin{equation} \lim _{x \rightarrow \infty} x^{2}\left(e^{-1 / x^{2}}-1\right) \end{equation}

Step-by-Step Solution

Answer

The limit is -1.

Step 1: Rewrite the Expression

The given limit is \( \lim_{x \to \infty} x^2 (e^{-1/x^2} - 1) \). We can rewrite this as \( \lim_{x \to \infty} x^2 (f(x) - 1) \), where \( f(x) = e^{-1/x^2} \). Our goal is to simplify and evaluate this limit.

Step 2: Expand Using Taylor Series

The Taylor series expansion for \( e^u \) around \( u=0 \) is \( e^u = 1 + u + \frac{u^2}{2} + \frac{u^3}{6} + \cdots \). For our function \( f(x) = e^{-1/x^2} \), let \( u = -1/x^2 \). Substituting into the series gives \( f(x) = 1 - \frac{1}{x^2} + \frac{1}{2x^4} - \frac{1}{6x^6} + \cdots \).

Step 3: Substitute the Expansion into the Limit

Substitute the expansion of \( e^{-1/x^2} \) into the limit expression: \[ \lim_{x \to \infty} x^2 \left( \left(1 - \frac{1}{x^2} + \frac{1}{2x^4} - \cdots \right) - 1 \right). \] This simplifies to \( \lim_{x \to \infty} x^2 \left( -\frac{1}{x^2} + \frac{1}{2x^4} - \cdots \right) \).

Step 4: Simplify and Evaluate the Limit

Multiply \( x^2 \) through the series: \( \lim_{x \to \infty} \left( -1 + \frac{1}{2x^2} - \cdots \right) \). As \( x \to \infty \), all other terms involving \( x \) in the denominator approach zero. So, the limit becomes \( -1 \).

Key Concepts

Taylor Series

The Taylor series is a powerful mathematical tool used to approximate functions by expressing them as an infinite sum of terms. Each term is derived from the derivatives of the function evaluated at a particular point. This concept is crucial when dealing with functions that are complex or difficult to handle in their standard form.

Here's how the Taylor series works:

• It takes a function that is infinitely differentiable at a point and expresses it as a power series.
• The expansion is centered at a specific point, often zero, which simplifies calculations.
• Its general form is given by: \[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots \] where \( f^n(a) \) is the \( n \)-th derivative of \( f \) evaluated at \( a \).
• This series allows for complex calculations such as finding the limits or solving differential equations.

In our case, we used the Taylor series to expand the exponential function \( e^{u} \). By setting \( u = -1/x^2 \), we simplified the otherwise challenging limit problem.

Series Expansion

Series expansion involves expressing functions as a sum of simpler terms. This method is advantageous because it can convert complex expressions into more manageable ones. When a function is expanded in this way, it becomes easier to analyze, especially as it relates to limits or continuity.

Some essential points about series expansion include:

• It is very useful in approximating functions that are hard to compute directly.
• The more terms included in the series, the more accurate the representation of the original function.
• Series can be finite, capturing only a part of the function, or infinite, which provides an exact representation if convergent.
• An example is the Taylor series, which is a specific type of series expansion centered at a particular point.

In the original exercise, the challenge was to evaluate the limit of a function with complex behavior as \( x \to \infty \). By expanding \( e^{-1/x^2} \) into its Taylor series, we were able to simplify the limit expression significantly, resulting in an approachable solution.

Exponential Function

The exponential function \( e^x \) is one of the most important functions in mathematics, often appearing in growth models, calculus, and physics, among others. Its defining characteristic is that the function's rate of change is proportional to its current value.

Understanding the exponential function better:

• The base of the exponential function is "\( e \)", which is approximately equal to 2.71828.
• It is defined as: \[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \] This is its Taylor series expansion about zero, showcasing its infinite series nature.
• In calculus, \( e^x \) is unique because its derivative and integral are both equal to \( e^x \).
• Exponential functions model continuous growth processes, such as population growth or compound interest.

In the exercise, the function \( e^{-1/x^2} \) was manipulated by a Taylor series expansion to simplify the evaluation of a limit. The intrinsic properties of exponential functions make them flexible and powerful tools in calculus and analysis.