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Find the limit. Use l'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. If l'Hospital's Rule doesn't apply, explain why. \(\lim _{x \rightarrow \infty} x \tan (1 / x)\)

Short Answer

Expert verified
The limit is 1.

Step by step solution

01

Identify the Expression Type

The expression is \(\lim_{x \to \infty} x \tan\left(\frac{1}{x}\right)\). As \(x\) approaches infinity, \(\frac{1}{x}\) approaches 0. We need to investigate the behavior of \(x \tan\left(\frac{1}{x}\right)\) as \(x\) grows larger.
02

Simplify the Expression

Start by substituting \(y = \frac{1}{x}\). As \(x \to \infty\), we have \(y \to 0^+\). Therefore, the limit becomes \(\lim_{y \to 0^+} \frac{\tan(y)}{y}\).
03

Evaluate the Simplified Limit

The expression \(\lim_{y \to 0^+} \frac{\tan(y)}{y}\) is a standard trigonometric limit known to equal 1. It can also be evaluated using l'Hospital's Rule, since \(\tan(y)\) and \(y\) both approach 0, forming an \(\frac{0}{0}\) indeterminate form. Differentiating the numerator and the denominator gives \(\lim_{y \to 0^+} \frac{\sec^2(y)}{1} = 1\).
04

Conclude the Original Limit

Since we transformed the original limit \(\lim_{x \to \infty} x \tan\left(\frac{1}{x}\right)\) into \(\lim_{y \to 0^+} \frac{\tan(y)}{y}\) and found it to be 1, the original limit is also 1.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Limit Evaluation
When dealing with limits, especially as a variable approaches infinity or zero, we are seeking to understand the behavior of a function as it gets close to a particular point. In the exercise provided, the goal is to evaluate the limit \[ \lim_{x \to \infty} x \tan \left(\frac{1}{x}\right) \]Here, as \(x\) tends towards infinity, \(\frac{1}{x}\) approaches zero. To simplify this, a substitution is used: let \(y = \frac{1}{x}\), making our limit:\[ \lim_{y \to 0^+} \frac{\tan(y)}{y} \]This approach simplifies the evaluation of the limit because it transforms the behavior of the function into a more familiar form—the limit of \( \frac{\tan(y)}{y} \) as \(y\) approaches zero, which is a classic trigonometric limit.
Indeterminate Forms
Indeterminate forms occur in calculus when evaluating limits that initially appear unresolved or undefined. The common types are \(\frac{0}{0}\) and \(\frac{\infty}{\infty}\). In our exercise, the expression \( \frac{\tan(y)}{y} \) as \(y \to 0\) forms an indeterminate type of \(\frac{0}{0}\), because both the numerator \(\tan(y)\) and the denominator \(y\) approach zero.To resolve this, we apply l'Hospital's Rule, which is used specifically for these types of indeterminate forms. By differentiating the numerator and the denominator separately:
  • The derivative of \(\tan(y)\) is \(\sec^2(y)\)
  • The derivative of \(y\) is 1
Hence, our limit becomes:\[ \lim_{y \to 0^+} \frac{\sec^2(y)}{1} = \lim_{y \to 0^+} \sec^2(y) = 1 \]Indeterminate forms require special techniques like l'Hospital's Rule to resolve them and find meaningful results.
Trigonometric Limits
Trigonometric limits often involve angles approaching zero and need particular attention due to their behavior near these points. The limit \( \lim_{y \to 0^+} \frac{\tan(y)}{y} \) is a fundamental trigonometric limit.This limit is significant because as \(y\) approaches zero, \(\tan(y)\) can be approximated by \(y\) itself since the two functions are very close in value. Therefore, \( \lim_{y \to 0^+} \frac{\tan(y)}{y} = 1 \) is a well-known result.Such trigonometric limits are crucial in calculus since they frequently simplify otherwise complex expressions and are essential to understanding the nature of trigonometric functions at small angles. Knowing these limits inside and out is a valuable tool when tackling calculus problems.

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