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Rewrite the indeterminate form of type \(0 \cdot \infty\) as either type \(\frac{0}{0}\) or type \(\frac{\infty}{\infty} .\) Use L'Hôpital's Rule to evaluate the limit. $$ \lim _{x \rightarrow \infty} 3\left(0.6^{x}\right)(\ln x) $$

Short Answer

Expert verified
The limit is 0.

Step by step solution

01

Identify Indeterminate Form

Examine the given limit expression \( \lim _{x \rightarrow \infty} 3(0.6^{x})(\ln x) \). As \( x \to \infty \), \( 0.6^{x} \to 0 \) and \( \ln x \to \infty \). Therefore, the expression is of the indeterminate form \( 0 \cdot \infty \).
02

Rewrite to Fraction Form

To apply L'Hôpital's Rule, rewrite the expression in a fraction form that results in either \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). Rewrite the expression as \[ \lim _{x \rightarrow \infty} \frac{\ln x}{(1/3)(0.6)^{-x}} \] so that it converts into \( \frac{\infty}{\infty} \) form since \( (0.6)^{-x} \to \infty \) as \( x \to \infty \).
03

Apply L'Hôpital's Rule

Since the expression is now in \( \frac{\infty}{\infty} \) form, apply L'Hôpital's Rule. Differentiate the numerator and the denominator with respect to \( x \); the derivative of \( \ln x \) is \( \frac{1}{x} \), and the derivative of \( (1/3)(0.6)^{-x} \) is \( \ln(0.6)(0.6)^{-x}/3 \).
04

Solve the New Limit

Solve the limit using L'Hôpital's Rule: \[ \lim_{x \to \infty} \frac{\frac{1}{x}}{-\frac{\ln(0.6)}{3} (0.6)^{-x}} = \lim_{x \to \infty} \frac{-3}{x (\ln(0.6)) (0.6)^{-x}} \]. Simplifying this further, and considering \( (0.6)^{-x} \to \infty \), leads to the conclusion that the limit equals 0.

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Key Concepts

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

Understanding Indeterminate Forms
In mathematics, especially in calculus, an **indeterminate form** arises when attempting to evaluate a limit and the result is not immediately clear. These forms arise from taking limits that involve operations like division or multiplication where initial results seem contradictory or undefined. Common indeterminate forms includegenerally:
  • \(0/0\)
  • \(\infty/\infty\)
  • \(0\cdot\infty\)
  • \(\infty - \infty\)
When you come across these forms, it means that there's more work to be done to simplify or rearrange the expression so that you can find the limit.
Typically, these forms signal that further algebraic manipulation or use of specific calculus rules, like L'Hôpital's Rule, is needed to resolve the limit. In the exercise, the form was initially \(0 \cdot \infty\). By rewriting it, we transformed it into \(\frac{\infty}{\infty}\), making it possible to apply L'Hôpital's Rule.
Mastering Limit Evaluation with L'Hôpital's Rule
**Limit evaluation** becomes interesting when the limit of a function as \(x\) approaches a certain value results in an indeterminate form. One powerful tool for evaluating such limits is L'Hôpital's Rule.
L'Hôpital's Rule helps calculate limits of indeterminate forms \(\frac{0}{0}\) and \(\frac{\infty}{\infty}\). When using this rule:
  • Differentiate the numerator and denominator separately.
  • Evaluate the new limit of the resulting derivative form.
Keep applying the rule as necessary until an evaluable limit appears.
In the exercise, once the limit was rewritten to \(\frac{\ln x}{(1/3)(0.6)^{-x}}\), it fell into the \(\frac{\infty}{\infty}\) category. Differentiating both the numerator and denominator transformed the expression, paving the way to finding that the limit equals zero.
The Role of Calculus in Solving Limits
**Calculus** is not just the study of change but also deeply involves the analysis and evaluation of limits. Limits are fundamental to calculus and are essential in understanding the behavior of functions as they approach particular points or infinity.
In solving limits, calculus provides tools like L'Hôpital's Rule, which specifically targets difficult forms like indeterminate ones. Without these tools, finding limits would be challenging, if not impossible, for complex expressions.
The exercise highlights how calculus completes the picture of a limit that, at first glance, seems unsolvable. By rearranging the expression and applying derivatives methodically, calculus breaks down what seems like unsolvable infinity interactions into manageable steps, ultimately revealing precise outcomes like the limit evaluated as 0.

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