Problem 12 - Calculus for Biology and Medicine

Question

Find the limits in Problems 1-60; not all limits require use of l'Hôpital's rule. $$ \lim _{x \rightarrow \infty} \frac{\ln x}{\sqrt{x}} $$

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Identify the Limit Type

Evaluate the expression to determine the indeterminate form. As \( x \to \infty \), both \( \ln x \) and \( \sqrt{x} \) tend towards infinity. Thus, the limit is of the indeterminate form \( \frac{\infty}{\infty} \).

Step 2: Apply l'Hôpital's Rule

Since the limit is in the \( \frac{\infty}{\infty} \) form, apply l'Hôpital's Rule, which states that \( \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \) if the limit exists. Differentiate the numerator and the denominator.

Step 3: Differentiate the Numerator and the Denominator

Differentiate \( \ln x \) to get \( \frac{1}{x} \) and differentiate \( \sqrt{x} \) to get \( \frac{1}{2\sqrt{x}} \).

Step 4: Simplify the New Expression

Substitute the derivatives into the limit: \[ \lim_{x \to \infty} \frac{\frac{1}{x}}{\frac{1}{2\sqrt{x}}} = \lim_{x \to \infty} \frac{2\sqrt{x}}{x} \]. Simplify this expression to \( \lim_{x \to \infty} \frac{2}{\sqrt{x}} \).

Step 5: Evaluate the Simplified Limit

As \( x \to \infty \), \( \sqrt{x} \to \infty \). Thus, \( \frac{2}{\sqrt{x}} \to 0 \). Therefore, the limit is \( 0 \).

Key Concepts

Indeterminate Forms

In calculus, when dealing with limits, you might stumble upon several special cases called indeterminate forms. These are expressions where the limit cannot be directly determined from the form alone, often appearing as basic, yet paradoxical ratios like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). For instance, our exercise \(\lim_{x \rightarrow \infty} \frac{\ln x}{\sqrt{x}}\) results in \(\frac{\infty}{\infty}\) as both the numerator, \(\ln x\), and denominator, \(\sqrt{x}\), grow infinitely large. Detecting these forms is crucial as they signal that direct substitution isn't sufficient, and instead, more advanced techniques such as l'Hôpital's Rule should be employed to find the limit. Recognizing and transforming indeterminate forms is a vital skill in calculus, allowing simplification of seemingly undefined expressions.

Differentiation Techniques

Differentiation is the mathematical process of finding the derivative, or rate of change, of a function. It's a foundational technique utilized in solving problems related to limits. In our original step-by-step solution, differentiation is applied to both functions, \(\ln x\) and \(\sqrt{x}\), to exploit l'Hôpital's Rule. The derivative of \(\ln x\) is \(\frac{1}{x}\), capturing the rate of change of the logarithmic function. Meanwhile, differentiating \(\sqrt{x}\) with respect to \(x\) yields the derivative \(\frac{1}{2\sqrt{x}}\). With these derivatives computed, we can re-evaluate the limit by substituting the derivatives in place of the original functions. Differentiation techniques, like the power rule and chain rule, often assist in converting challenging expressions into simpler forms for solving limits.

Limits at Infinity

Limits can describe the behavior of functions as the input values grow exceedingly large or small, often referred to as tending towards infinity. Understanding limits at infinity is essential for studying functions that approach particular values asymptotically as \(x\) approaches positive or negative infinity. The given exercise involves evaluating a limit at infinity, specifically \(\lim_{x \to \infty} \frac{2}{\sqrt{x}}\). Here, as \(x\) becomes very large, \(\sqrt{x}\) also becomes very large, leading \(\frac{2}{\sqrt{x}}\) to approach 0. Consequently, recognizing how functions behave at infinity is key to unlocking the behavior of an entire function in a particular direction. This helps identify horizontal asymptotes and the overall long-term behavior of functions. Applying the correct techniques ensures accurate interpretation of function behaviors as they extend infinitely.