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Find the limit, if it exists. $$ \lim _{x \rightarrow \infty} \frac{\ln x}{x^{2}} $$

Short Answer

Expert verified
The limit is 0.

Step by step solution

01

Identify the Infinity Form

As we analyze the expression \(\frac{\ln x}{x^{2}}\) for \(x \rightarrow \infty\), we notice that as \(x\) becomes very large, \(\ln x\) tends towards infinity and \(x^2\) also tends towards infinity. This creates an indeterminate form of \(\frac{\infty}{\infty}\).
02

Apply L'Hôpital's Rule

Since we have an indeterminate form, we apply L'Hôpital's Rule, which is applicable when direct substitution yields \(\frac{\infty}{\infty}\). L'Hôpital's Rule states that \(\lim_{x \to \infty} \frac{f(x)}{g(x)} = \lim_{x \to \infty} \frac{f'(x)}{g'(x)}\) if the limit on the right hand side exists. Here, \(f(x) = \ln x\) and \(g(x) = x^2\).
03

Differentiate the Numerator and Denominator

Differentiate the numerator \(\ln x\) to get \(f'(x) = \frac{1}{x}\). Differentiate the denominator \(x^2\) to get \(g'(x) = 2x\). The new limit becomes \(\lim_{x \to \infty} \frac{1/x}{2x}\).
04

Simplify the Resulting Expression

The expression \(\frac{1/x}{2x}\) can be rewritten as \(\frac{1}{2x^2}\).
05

Evaluate the Simplified Limit

Evaluate \(\lim_{x \to \infty} \frac{1}{2x^2}\). As \(x\) approaches infinity, the term \(2x^2\) becomes very large, making \(\frac{1}{2x^2}\) approach zero.
06

Conclude the Limit Calculation

Since the limit of \(\frac{1}{2x^2}\) as \(x\) approaches infinity is zero, it follows that the original limit \(\lim_{x \rightarrow \infty} \frac{\ln x}{x^{2}} = 0\). Therefore, the limit exists and equals zero.

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Key Concepts

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

Indeterminate Forms
When you first learn about limits, you'll often encounter expressions that seem difficult to evaluate directly. Some of these expressions fall into the category of "indeterminate forms." Indeterminate forms occur when substituting a particular value, often infinity, into a limit leads to undefined operations like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). These forms do not immediately reveal information about what the limit actually is.

Understanding indeterminate forms is crucial because they signal the need for a different approach to find the limit. In the original exercise, the expression \(\frac{\ln x}{x^2}\) as \(x \to \infty\) is indeterminate because both the numerator \(\ln x\) and the denominator \(x^2\) become infinite. Hence, the form \(\frac{\infty}{\infty}\) arises.

Recognizing these forms allows you to use techniques such as applying L'Hôpital's Rule, factoring, or other algebraic manipulations, to simplify the expression and find the actual limit.
Limit Calculation
Calculating limits, especially those involving indeterminate forms, often requires a bit of strategy. Instead of just plugging in values, you might need mathematical tools like L'Hôpital's Rule, which is designed specifically for dealing with indeterminate forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\).

In our exercise example, this rule is applied to the expression \(\lim_{x \to \infty} \frac{\ln x}{x^2}\). Here's the step-by-step breakdown:
  • Recognize the indeterminate form \(\frac{\infty}{\infty}\).
  • L'Hôpital's Rule states that for functions \(f(x)\) and \(g(x)\), if \(\lim_{x \to c} \frac{f(x)}{g(x)} = \frac{\infty}{\infty}\), then
    \[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \] provided this new limit exists.


Using this, you differentiate the numerator and denominator individually and re-calculate the limit with these new derivatives, often simplifying the process and revealing the true limit.
Differentiation
Differentiation is a fundamental tool in calculus, and it plays an essential role when applying L'Hôpital's Rule to indeterminate forms. In essence, differentiation involves finding the derivative of a function, which measures how the function value changes as its input changes.

In the context of our exercise, after recognizing the indeterminate form \(\frac{\infty}{\infty}\), the functions \(f(x) = \ln x\) and \(g(x) = x^2\) are differentiated to find their derivatives. This process involves:
  • For the numerator \(f(x) = \ln x\), the derivative \(f'(x)\) is \(\frac{1}{x}\).
  • For the denominator \(g(x) = x^2\), the derivative \(g'(x)\) is \(2x\).


This transformation turns the problematic limit into a simpler one: \(\lim_{x \to \infty} \frac{1/x}{2x}\). Evaluating this simplified form reveals the limit as zero, concluding the calculation effectively. This illustrates the power of differentiation in tackling complex calculus problems.

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