Question

Use the following argument to show that \(\lim _{x \rightarrow \infty} \ln x=\infty\) and \(\lim _{x \rightarrow 0^{+}} \ln x=-\infty\). a. Make a sketch of the function \(f(x)=1 / x\) on the interval \([1,2] .\) Explain why the area of the region bounded by \(y=f(x)\) and the \(x\) -axis on [1,2] is \(\ln 2\). b. Construct a rectangle over the interval [1,2] with height \(1 / 2\) Explain why \(\ln 2>1 / 2\). c. Show that \(\ln 2^{n}>n / 2\) and \(\ln 2^{-n}<-n / 2\). d. Conclude that \(\lim _{x \rightarrow \infty} \ln x=\infty\) and \(\lim _{x \rightarrow 0^{+}} \ln x=-\infty\).

Short answer

Question: Prove that \(\lim _{x \rightarrow \infty} \ln x=\infty\) and \(\lim _{x \rightarrow 0^{+}} \ln x=-\infty\). Solution: - We analyzed the function \(f(x) = 1/x\) and found that the area of the region bounded by \(y = f(x)\) on \([1, 2]\) is \(\ln 2\). - We constructed a rectangle with a height of \(1/2\) on the interval \([1, 2]\) and showed that its area is less than \(\ln 2\). - By applying inequalities, we showed that \(\ln 2^{n} > n/2\) and \(\ln 2^{-n} < -n/2\). - Finally, we concluded that \(\lim _{x \rightarrow \infty} \ln x=\infty\) and \(\lim _{x \rightarrow 0^{+}} \ln x=-\infty\). Thus, we have proven that the limits of the natural logarithm function are as required.

Step-by-step solution

Sketch and Area of the Region

To start, we will sketch the function \(f(x) = 1/x\) on the interval \([1, 2]\). The function \(f(x) = 1/x\) is a standard hyperbola with an asymptote at the x-axis and the y-axis. The shape of the function is a curved downward-sloping line in the first quadrant. Now, we need to explain why the area of the region bounded by \(y = f(x)\) and the x-axis on \([1, 2]\) is \(\ln 2\). To do this, we'll consider the improper integral \(\int_{1}^{2} f(x) dx\). We have: $$\int_{1}^{2} \frac{1}{x} dx = \int_{1}^{2} \ln x' dx$$ Taking the antiderivative of the natural logarithm, we get $$[\ln x]_{1}^{2} = \ln 2 - \ln 1 = \ln 2$$ So, the region bounded by the curve \(y = f(x)\) and the x-axis on \([1, 2]\) has an area equal to \(\ln 2\).

Construct a Rectangle

Next, we will construct a rectangle over the interval \([1, 2]\) with height \(1/2\). The base of the rectangle is \(2 - 1 = 1\). Thus, the area of the rectangle is $$A_{rectangle} = \text{base} \times \text{height} = 1 \times \frac{1}{2} = \frac{1}{2}$$ Since the function \(f(x) = 1/x\) is positive and strictly decreasing for \(x > 0\), we know that this rectangle is fully contained within the region bounded by \(y = f(x)\) and the x-axis on \([1, 2]\). Hence, the area of this region is greater than the area of the rectangle, meaning $$\ln 2 > \frac{1}{2}$$

Show Inequalities for \(\ln 2^{n}\) and \(\ln 2^{-n}\)

Now, we need to show that \(\ln 2^{n} > n/2\) and \(\ln 2^{-n} < -n/2\). Using the logarithmic property of exponentiation, we can rewrite both expressions: $$\ln 2^{n} = n \ln 2 \quad \text{and} \quad \ln 2^{-n} = -n \ln 2$$ We have already shown that \(\ln 2 > 1/2\). Therefore, by multiplying both sides of the inequality by \(n\), which is a positive number, we get $$n \ln 2 > n \frac{1}{2}$$ And, by multiplying both sides of the inequality by \(-n\), which is a negative number, we get $$-n \ln 2 < -n \frac{1}{2}$$ Thus, we have proven the required inequalities: $$\ln 2^n > \frac{n}{2} \quad \text{and} \quad \ln 2^{-n} < -\frac{n}{2}$$

Conclude the Limits

Finally, we can conclude that \(\lim _{x \rightarrow \infty} \ln x=\infty\) and \(\lim _{x \rightarrow 0^{+}} \ln x=-\infty\). For the first limit, as \(x \rightarrow \infty\), consider \(x = 2^n\). Since we have shown \(\ln 2^n > n/2\), we have \(\ln x > n/2\). As \(n\) approaches infinity, so does \(x\). Therefore, \(\ln x\) will also approach infinity, resulting in $$\lim _{x \rightarrow \infty} \ln x=\infty$$ For the second limit, as \(x \rightarrow 0^{+}\), consider \(x = 2^{-n}\). Since we have shown \(\ln 2^{-n} < -n/2\), we have \(\ln x < -n/2\). As \(n\) approaches infinity, \(x\) approaches zero from the right. Therefore, \(\ln x\) will approach \(-\infty\), resulting in $$\lim _{x \rightarrow 0^{+}} \ln x=-\infty$$ Thus, we have proven the required limits using the arguments provided by the exercise.

Key concepts

Improper Integrals

Improper integrals may sound complicated, but they are simply a type of integral involving unbounded limits or unbounded functions. For example, when we integrate the function \(f(x) = 1/x\) along the interval \([1, \infty)\), we are dealing with an improper integral because it extends to infinity.
To evaluate an improper integral, we often take the limit of another integral. For instance, we can rewrite\[ \int_{1}^{\infty} \frac{1}{x} \, dx = \lim_{b \to \infty} \int_{1}^{b} \frac{1}{x} \, dx. \] This allows us to perform usual integration on the interval \([1, b]\) and then analyze what happens as \(b\) approaches infinity. Through this process, we find out if the integral converges (has a finite value) or diverges (grows without bound).

• Improper integrals involve infinite limits or discontinuous functions.
• They require the use of limits to evaluate.
• They help us understand areas under curves that stretch infinitely or behave wildly at certain points.

Exploring improper integrals can be a great way to understand the limits of functions like logarithmic ones, especially when determining their behavior as they go to infinity or approach zero.

Logarithmic Inequalities

Logarithmic inequalities help us compare the sizes of numbers in terms of their logarithms. Consider the function \(f(x) = \ln(x)\), which is the natural logarithm of \(x\). Since this function is increasing, smaller \(x\) values lead to smaller logarithmic values, and larger \(x\) values lead to larger logarithmic values.
Let's look at an example: we want to prove that \(\ln 2 > \frac{1}{2}\). By using a geometric illustration or constructing rectangles, like we did in the exercise, we establish the idea visually. The area under the curve \(f(x) = 1/x\) from \(1\) to \(2\) is \(\ln 2\), which proves to be larger than the area of a rectangle under the same interval with height \(1/2\).

• Logarithmic inequalities allow comparisons using logarithmic expressions.
• The natural logarithm function \(\ln(x)\) is monotonically increasing.
• These inequalities are useful for assessing growth and decay rates in logarithms.

By gaining a grasp of logarithmic inequalities, we develop a deeper understanding of asymptotic behavior in functions.

Asymptotic Behavior

Asymptotic behavior describes how functions behave as their input approaches certain critical values, like infinity or zero. For logarithmic functions like \(\ln(x)\), there's a unique asymptotic behavior.
Consider that \(\lim_{x \rightarrow \infty} \ln x = \infty\), meaning as \(x\) grows larger, the logarithm grows without bound. Conversely, \(\lim_{x \rightarrow 0^{+}} \ln x = -\infty\) signifies that as \(x\) moves closer to zero from the positive side, the logarithmic value dives to negative infinity. These trends are a fundamental aspect of how logarithmic functions interact with numbers, illustrating exponential-like growth in reverse

• Asymptotic behavior involves limits as inputs grow very large or tiny.
• Logarithmic functions grow slowly but continue indefinitely.
• Understanding their limits enhances comprehension of dynamic limits in calculus.

By knowing the asymptotic behavior of logarithms, we can predict and explain the fate of various mathematical models and equations as their variables stretch towards infinite boundaries.