Question

Use analytical methods to evaluate the following limits. $$\lim _{n \rightarrow \infty}\left(\cot \frac{1}{n}-n\right)$$

Short answer

Answer: The limit of the expression \(\lim _{n \rightarrow \infty}\left(\cot \frac{1}{n} - n\right)\) is -1.

Step-by-step solution

Simplify the expression

We rewrite the expression as: $$\lim _{n \rightarrow \infty}\left(\cot \frac{1}{n}-n\right) = \lim _{n \rightarrow \infty}\left[\frac{\cos(\frac{1}{n})}{\sin(\frac{1}{n})} - n\right]$$

Make the expression suitable for L'Hôpital's Rule

We should rewrite the expression in terms of division so that we can apply L'Hôpital's Rule later. To do that, set \(x = \frac{1}{n}\) and as n goes to infinity, x will approach \(0\). The expression then becomes: $$\lim _{x \rightarrow 0}\left[\frac{\cos x}{x\,\sin x}\right] - \lim _{x \rightarrow 0}\frac{1}{x}$$

Apply L'Hôpital's Rule

The first part of the expression, \(\lim _{x \rightarrow 0}\left[\frac{\cos x}{x\,\sin x}\right]\), can be evaluated using L'Hôpital's Rule which requires the expression to be in "0/0" or "infinity/infinity" form. Thus, we apply L'Hôpital's Rule, by taking the derivative of both numerator and denominator with respect to x: $$\lim _{x \rightarrow 0}\frac{\cos x}{x\,\sin x} = \lim _{x \rightarrow 0}\frac{-\sin x}{x\,\cos x + \sin x}$$ Since the result is again in the "0/0" form, we apply L'Hôpital's Rule once more: $$\lim _{x \rightarrow 0}\frac{-\sin x}{x\,\cos x + \sin x} = \lim _{x \rightarrow 0}\frac{-\cos x}{-\sin x + \cos x -x\,\sin x}$$ Now, we evaluate the second part of the expression, \(\lim _{x \rightarrow 0}\frac{1}{x}\). Since this expression is in the "infinity/infinity" form: $$\lim _{x \rightarrow 0}\frac{1}{x} = \lim_{x \rightarrow 0} \frac{0}{1} = 0$$

Find the complete limit

Now we can write the complete limit expression, adding the results together: $$\lim _{n \rightarrow \infty}\left(\cot \frac{1}{n} - n\right) = \lim _{x \rightarrow 0}\frac{-\cos x}{-\sin x + \cos x -x\,\sin x} + 0$$ As x approaches 0, the expression becomes: $$\lim _{x \rightarrow 0}\frac{-\cos x}{-\sin x + \cos x -x\,\sin x} = \frac{-\cos(0)}{-\sin(0) + \cos(0) - 0\cdot \sin(0)} = \frac{-1}{1}=-1$$ Therefore, the limit of the expression is: $$\lim _{n \rightarrow \infty}\left(\cot \frac{1}{n} - n\right) = -1$$

Key concepts

L'Hôpital's Rule

L'Hôpital's Rule is a vital tool in calculus that helps us solve limit problems involving indeterminate forms like \(0/0\) or \(\infty/\infty\). It simplifies complex limits by taking derivatives to evaluate the behavior of functions at specific points, usually close to zero or infinity. When confronted with such forms, L'Hôpital's Rule enables us to:

• Take the derivative of the numerator and the derivative of the denominator separately.
• Re-evaluate the limit after differentiating.

Let's consider a situation where we're dealing with a limit such as \( \lim_{x \to c} \frac{f(x)}{g(x)}\). If both the numerator and denominator approach zero or infinity as \(x\) approaches \(c\), we can apply L'Hôpital's Rule:

• Compute \( \lim_{x \to c} \frac{f'(x)}{g'(x)} \).
• If this limit is still an indeterminate form, you may need to apply the rule iteratively.

L'Hôpital's Rule helps simplify the evaluation of challenging limits, as shown in the exercise solution.

Cotangent Function

The cotangent function, denoted as \( \cot(x) \), is the reciprocal of the tangent function. The cotangent of an angle \(x\) in a right triangle is defined as the ratio of the adjacent side to the opposite side. In terms of trigonometric functions, it can be expressed as:

• \( \cot(x) = \frac{1}{\tan(x)} = \frac{\cos(x)}{\sin(x)} \)

In the context of limits and calculus, the cotangent function sometimes poses challenges due to its behavior near zeros of the sine function. As \(\sin(x)\) approaches zero, \( \cot(x) \) may approach infinity or negative infinity based on its direction. This exercise involves:

• Expressing \( \cot\left(\frac{1}{n}\right) \) which can be rewritten as \( \frac{\cos\left(\frac{1}{n}\right)}{\sin\left(\frac{1}{n}\right)} \).
• Understanding how these ratios behave as \( n \to \infty \).

Understanding the behavior of \( \cot(x) \) is crucial when evaluating limits, especially when approaching infinity or zero.

Infinity Limits

Infinity limits are concerned with what happens to a function as the input approaches infinity or as the function itself grows indefinitely. These limits help us understand the long-term behavior of functions in calculus. When dealing with infinity limits, you should apply certain strategies:

• Identify if the function is growing without bound or if it's approaching a finite value.
• Use simplification techniques like factoring or division by the highest power for polynomial ratios.
• Apply rules like L'Hôpital's Rule for indeterminate forms to find limits at infinity.

In the problem we solved, \(\lim _{n \rightarrow \infty}\left(\cot \frac{1}{n} - n\right)\), we are tasked to observe how the function changes as \( n \to \infty \). Since \( \cot\left(\frac{1}{n}\right) \) translates to \( \frac{\cos\left(\frac{1}{n}\right)}{\sin\left(\frac{1}{n}\right)} \), understanding how basic trigonometric functions behave at infinity becomes critical.Evaluating limits at infinity reveals more than just a value; it highlights the potential growth direction or stability of a function as its inputs get larger. This is particularly important in infinite context problems.