Problem 20 - Calculus Early Transcendentals

Question

Evaluate the following limits using \(l\) Hópital's Rule. $$\lim _{x \rightarrow 2 \pi} \frac{x \sin x+x^{2}-4 \pi^{2}}{x-2 \pi}$$

Step-by-Step Solution

Answer

Question: Evaluate the limit using L'Hôpital's Rule: \(\lim_{x\rightarrow 2\pi} \frac{x\sin x + x^2 - 4\pi^2}{x-2\pi}\) Answer: \(\lim_{x\rightarrow 2\pi} \frac{x\sin x + x^2 - 4\pi^2}{x-2\pi} = 6\pi\)

Step 1: Determine if we can apply L'Hôpital's Rule

Let's substitute the limit value, \(x = 2\pi\), into the given function. $$\frac{(2\pi) \sin (2\pi) + (2\pi)^2 - 4\pi^2}{(2\pi) - 2\pi} = \frac{0}{0}$$ Since the given function is in the indeterminate form of 0/0, we can apply L'Hôpital's Rule.

Step 2: Apply L'Hôpital's Rule

Using L'Hôpital's Rule, we will find the derivative of both the numerator and the denominator. $$\frac{d}{dx}\left(x\sin x + x^2 - 4\pi^2\right) = \sin x + x\cos x + 2x$$ $$\frac{d}{dx}\left(x -2\pi\right) = 1$$ Now, we will find the limit of the ratio of the derivatives. $$\lim_{x\rightarrow 2\pi} \frac{\sin x + x\cos x + 2x}{1}$$

Step 3: Evaluate the limit

As \(x \rightarrow 2\pi\), the sine and cosine functions will have clear values, and we won't need L'Hôpital's Rule again. $$\lim_{x\rightarrow 2\pi}(\sin x + x\cos x + 2x) = \sin(2\pi) + 2\pi\cos(2\pi) + 2(2\pi)$$ $$ = 0 + 2\pi(1) + 4\pi = 2\pi + 4\pi$$ So the limit is: $$\lim_{x\rightarrow 2\pi} \frac{x\sin x + x^2 - 4\pi^2}{x-2\pi} = 6\pi$$

Key Concepts

Limits and Continuity

The concept of limits is fundamental to calculus and, more broadly, mathematical analysis. Limits describe the behavior of a function as its argument approaches a certain point. In the example given, we're asked to evaluate a limit as the variable approaches the value of \(2\pi\).
Continuity, on the other hand, is a property of a function if it does not have any abrupt changes, holes, or jumps in its graph. For a function to be continuous at a particular point, several conditions must be met: the function must be defined at that point, the limit as the variable approaches that point must exist, and the value of the limit must equal the value of the function at that point.
Understanding limits and continuity is crucial when dealing with L'Hôpital's Rule because this rule can only be applied to functions that are differentiable in the neighborhood of the point of interest, except possibly at the point itself.

Derivatives Calculus

In calculus, the derivative of a function represents the rate at which the function's value is changing at a particular point. Formally, the derivative at a point is the limit of the function's average rate of change in a tiny interval around that point. When we differentiate a function, we're essentially calculating this limit.
To solve the given limit using L'Hôpital's Rule, we first needed to find the derivatives of the numerator and the denominator. The derivative operation applied to \(x\sin x + x^2 - 4\pi^2\) gives us a new function, which in the limit yields \(2\pi + 4\pi\). Derivatives are a powerful tool because they allow us to transform indeterminate forms like \(0/0\) into values that we can evaluate directly.

Indeterminate Form

In the context of limits, an indeterminate form such as \(0/0\) arises when both the numerator and denominator of a fraction approach zero as the variable approaches a certain point. At first glance, these forms do not offer clear information on what the behavior of the function is at the limit point. Indeterminate forms can also include other types such as \(\infty/\infty\), \(0\cdot\infty\), \(\infty - \infty\), \(^{0}0\), \(^{\infty}0\), and more.
L'Hôpital's Rule is specifically designed to resolve these indeterminate forms by differentiating the numerator and the denominator separately and then taking the limit of the resulting function. This allows us to bypass the indeterminate form and find a concrete value for the limit, as was done in the example to find that the limit as \(x\rightarrow 2\pi\) equaled \({6\pi\).