Problem 56 - Calculus Early Transcendentals

Question

Using L'Hôpital's rule (Section 3.6) one can verify that $$ \lim _{x \rightarrow+\infty} \frac{\ln x}{x^{r}}=0, \quad \lim _{x \rightarrow+\infty} \frac{x^{r}}{\ln x}=+\infty, \quad \lim _{x \rightarrow 0^{+}} x^{r} \ln x=0 $$ for any positive real number \(r\). In these exercises: (a) Use these results, as necessary, to find the limits of \(f(x)\) as \(x \rightarrow+\infty\) and as \(x \rightarrow 0^{+} .\) (b) Sketch a graph of \(f(x)\) and identify all relative extrema, inflection points, and asymptotes (as appropriate). Check your work with a graphing utility. Using L'Hôpital's rule (Section 3.6) one can verify that $$ \lim _{x \rightarrow+\infty} \frac{\ln x}{x^{r}}=0, \quad \lim _{x \rightarrow+\infty} \frac{x^{r}}{\ln x}=+\infty, \quad \lim _{x \rightarrow 0^{+}} x^{r} \ln x=0 $$ for any positive real number \(r\). In these exercises: (a) Use these results, as necessary, to find the limits of \(f(x)\) as \(x \rightarrow+\infty\) and as \(x \rightarrow 0^{+} .\) (b) Sketch a graph of \(f(x)\) and identify all relative extrema, inflection points, and asymptotes (as appropriate). Check your work with a graphing utility. $$ f(x)=x^{2} \ln x $$

Step-by-Step Solution

Answer

The limits are \(0\) as \(x \to +\infty\) and \(x \to 0^+\). The function has a relative extremum at \(x = \frac{1}{e}\) and an inflection point at \(x = \frac{1}{\sqrt{e^3}}\). No asymptotes.

Step 1: Identify the Limits Problem

We are given the function \(f(x) = x^2 \ln x\). We need to find the limit of this function as \(x \rightarrow +\infty\) and \(x \rightarrow 0^{+}\). We are told that for any positive real number \(r\), certain limit results apply.

Step 2: Apply Limit as x approaches Infinity

Given \(\lim_{x \to +\infty} \frac{\ln x}{x^r} = 0\), we use substitution to express \(f(x)\) as \(\frac{\ln x}{x^{-2}}\) when \(x \rightarrow +\infty\). Here, \(r = 2\). Hence, \(\lim_{x \to +\infty} \frac{\ln x}{x^{-2}} = 0\) implies that \(\lim_{x \to +\infty} x^2 \ln x = 0\).

Step 3: Apply Limit as x approaches Zero from the Right

For \(\lim_{x \to 0^+} x^r \ln x = 0\), we need to match this with \(f(x) = x^2 \ln x\), substituting \(r = 2\). Thus, \(\lim_{x \to 0^+} x^2 \ln x = 0\) directly from the given information.

Step 4: Analyze the Derivative to Find Relative Extrema

Find the derivative of \(f(x) = x^2 \ln x\) to determine critical points. Using the product rule, the derivative is \(f'(x) = 2x \ln x + x\). Setting \(f'(x) = 0\) solves to find critical points where \(x = \frac{1}{e}\), indicating a relative extremum.

Step 5: Analyze the Second Derivative for Concavity and Inflection Points

Find the second derivative \(f''(x) = 2\ln x + 3\). Setting \(f''(x) = 0\) identifies potential inflection points. Solving leads to \(\ln x = -\frac{3}{2}\), or \(x = \frac{1}{\sqrt{e^3}}\), showing a change in concavity at this point.

Step 6: Identify Asymptotes

As \(x \rightarrow +\infty\), \(\lim_{x \to +\infty} x^2 \ln x = 0\) establishes no vertical or slant asymptotes. As \(x \rightarrow 0^+\), \(x^2 \ln x \rightarrow 0\). No horizontal asymptotes exist since the limits are zero at both extremes.

Step 7: Graph Sketching and Utility Check

Sketch \(f(x) = x^2 \ln x\) marking the relative extremum at \(x = \frac{1}{e}\) and an inflection point at \(x = \frac{1}{\sqrt{e^3}}\). Check the graph using a graphing utility to validate these points of behavior.

Key Concepts

Limit of Logarithmic Functions

Understanding limits is crucial in calculus, and the limit of logarithmic functions like \( \ln x \) is particularly important. When dealing with the limit \( \lim_{x \to +\infty} \frac{\ln x}{x^r} = 0 \), you encounter scenarios where the numerator \( \ln x \) grows slowly compared to \( x^r \), making the fraction approach zero as \( x \) becomes infinitely large.

For functions involving products like \( x^2 \ln x \), this principle helps us understand their behavior at the extremes. Similarly, the expression \( \lim_{x \to 0^+} x^r \ln x = 0 \) is vital when \( x \) approaches zero from the positive side, where the product still approaches zero. This is because \( \ln x \) goes to negative infinity as \( x \to 0^+ \), but is counteracted by \( x^r \) quickly approaching zero, making their product diminish to zero.

Key things to remember:

• \( \ln x \) grows slower than any power of \( x \).
• Near zero, \( \ln x \) becomes large and negative, but multiplied with small \( x^r \) it approaches zero.

Relative Extrema

Relative extrema are the points on a graph where a function changes direction from increasing to decreasing or vice versa. They are found using the derivative of the function, which tells us about the slope of the curve.

For the function \( f(x) = x^2 \ln x \), the derivative \( f'(x) = 2x \ln x + x \) helps identify these points. Setting \( f'(x) = 0 \), you solve for \( x \) to find critical points. In this case, the critical point at \( x = \frac{1}{e} \) indicates a relative extremum because the slope of the curve changes from positive to negative or vice versa.

To determine whether it’s a maximum or minimum:

• Analyze where \( f'(x) > 0 \) and \( f'(x) < 0 \).
• The sign of \( f'(x) \) tells whether the function is increasing or decreasing.

Inflection Points

Inflection points are where the curve changes concavity, either from concave up to concave down, or vice versa.

We find inflection points by analyzing the second derivative \( f''(x) \). For \( f(x) = x^2 \ln x \), the second derivative \( f''(x) = 2\ln x + 3 \) helps us identify these changes. Setting \( f''(x) = 0 \), you find that \( x = \frac{1}{\sqrt{e^3}} \), indicating a potential inflection point.

To confirm a true inflection point:

• Check the sign change of \( f''(x) \) around the point.
• If \( f''(x) \) changes signs, the concavity changes, confirming it as an inflection point.

Asymptotes

Asymptotes give us insight into the behavior of functions at extreme values of \( x \). Often, they appear as lines that the graph approaches but never quite reaches.

For \( f(x) = x^2 \ln x \), there are no vertical or slant asymptotes as \( x \to +\infty \) since \( \lim_{x \to +\infty} x^2 \ln x = 0 \). Similarly, \( x^2 \ln x \to 0 \) as \( x \to 0^+ \).

Key notes:

• There are no horizontal asymptotes in this case.
• As \( x \to +\infty \), the function does not shoot off to infinity but stabilizes at zero.