Question

ln x is unbounded Use the following argument to show that lim (x → ∞) ln x = ∞ and lim (x → 0⁺) ln x = −∞.
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.

Accepted Answer

Step 1: Recall that the natural logarithm function $$ \ln x $$ can be defined as the integral $$ \ln x = \int_1^x \frac{1}{t} \, dt . $$This means the value of $$ \ln x $$ represents the area under the curve $$ y = \frac{1}{t} $$ from $$ t = 1  to$$ $$ t = x . $$Step 2: Sketch the function $$ f(x) = \frac{1}{x}  on $$the interval $$[1, 2]. $$The curve starts at $$ f(1) = 1 $$ and decreases to $$ f(2) = \frac{1}{2} . $$The area under this curve between $$ x = 1 $$ and $$ x = 2 $$, bounded by the x-axis, corresponds exactly to $$ \int_1^2 \frac{1}{x} \, dx = \ln 2 . $$Step 3: Explain that since $$ \ln 2  is $$the area under $$ y = \frac{1}{x} $$ from 1 to 2, this integral interpretation helps us understand the behavior of $$ \ln x  as$$ $$ x $$ changes. The area grows without bound as $$ x 	o \infty $$, showing $$ \lim_{x 	o \infty} \ln x = \infty . $$Step 4: Similarly, consider the limit as $$ x 	o 0^+ . $$The integral $$ \int_x^1 \frac{1}{t} \, dt = -\ln x $$ represents the area under $$ y = \frac{1}{t} $$ from $$ t = x  to 1. As$$ $$ x $$ approaches 0 from the right, this area grows without bound, implying $$ \ln x 	o -\infty . $$Step 5: Summarize that the integral definition of $$ \ln x  as $$the area under $$ y = \frac{1}{x} $$ provides a clear geometric interpretation of why $$ \ln x  is $$unbounded: it increases without limit as $$ x 	o \infty $$ and decreases without limit as $$ x 	o 0^+ .$$

ln x is unbounded Use the following argument to show that lim (x → ∞) ln x = ∞ and lim (x → 0⁺) ln x = −∞.
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.

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Key Concepts

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ln x is unbounded Use the following argument to show that lim (x → ∞) ln x = ∞ and lim (x → 0⁺) ln x = −∞.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.

Verified video answer for a similar problem:

Key Concepts

Natural Logarithm as an Integral

Behavior of the Integral and Limits

Area Under a Curve and Definite Integrals

ln x is unbounded Use the following argument to show that lim (x → ∞) ln x = ∞ and lim (x → 0⁺) ln x = −∞.
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.