Problem 50 - Calculus Early Transcendentals

Question

Using L'Hôpital's rule (Section 3.6) one can verify that $$ \lim _{x \rightarrow+\infty} \frac{e^{x}}{x}=+\infty, \quad \lim _{x \rightarrow+\infty} \frac{x}{e^{x}}=0, \quad \lim _{x \rightarrow-\infty} x e^{x}=0 $$ In these exercises: (a) Use these results, as necessary, to find the limits of \(f(x)\) as \(x \rightarrow+\infty\) and as \(x \rightarrow-\infty\). (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. $$ f(x)=e^{-1 / x^{2}} $$

Step-by-Step Solution

Answer

As \( x \to \pm\infty, f(x)\to 1;\) as \( x \to 0, f(x) \to 0;\) with a horizontal asymptote at \( y = 1 \).

Step 1: Analyze the behavior of the function as x approaches +∞

For the function \( f(x) = e^{-1 / x^2} \), consider what happens as \( x \to +\infty \). The term \( -1 / x^2 \) approaches 0, thus giving \( e^0 = 1 \). Therefore, \( \lim_{x \to +\infty} f(x) = 1 \).

Step 2: Analyze the behavior of the function as x approaches -∞

Similarly, as \( x \to -\infty \), the expression \( -1 / x^2 \) also approaches 0 because the square of \( x \) makes it positive, returning \( e^0 = 1 \). Thus, \( \lim_{x \to -\infty} f(x) = 1 \).

Step 3: Determine vertical asymptotes and behavior around x = 0

The only potential difficulty point is \( x = 0 \), where \( f(x) = e^{-1/x^2} \) becomes undefined as \( x \to 0^+ \) or \( x \to 0^- \). As \( x \to 0^+ \), \( 1/x^2 \to +\infty \), thus \( e^{-1/x^2} \to 0 \). As \( x \to 0^- \), we have the same situation, \( f(x) \to 0 \).

Step 4: Identify asymptotes and sketching f(x)

Since \( f(x) \to 1 \) as \( x \to \pm\infty \) and \( f(x) \to 0 \) as \( x \to 0 \), there is a horizontal asymptote at \( y = 1 \) and the graph approaches the x-axis as \( x \to 0 \).

Step 5: Check for critical points and concavity

To find critical points and inflection points, one would take the first and second derivatives of \( f(x) \) and analyze them. In this function, the critical and inflection analysis shows no relative maxima or minima due to the structure of the exponential function being smooth between asymptotes.

Key Concepts

Limits of Functions

Limits are fundamental in calculus. They describe how a function behaves as the input approaches a particular value. For the function \( f(x) = e^{-1/x^2} \), evaluating limits gives insight into the function's behavior at the extremes.

• As \( x \to +\infty \), the term \(-1/x^2\) approaches 0, leading to \( e^{0} = 1 \). Hence, \( \lim_{x \to +\infty} f(x) = 1 \).
• Likewise, as \( x \to -\infty \), \(-1/x^2 \) again approaches 0, returning \( e^{0} = 1 \). Thus, \( \lim_{x \to -\infty} f(x) = 1 \).

Understanding limits helps in predicting the end behavior of functions, which is crucial for sketching graphs and analyzing outcomes.

Exponential Functions

Exponential functions are characterized by a constant base raised to a variable exponent. The function \( f(x) = e^{-1/x^2} \) exemplifies a more complex form.

• Here, the exponent includes a variable in the denominator, causing unique behavior.
• Exponential functions grow or decay rapidly; \( f(x) \) subtly approaches a limit due to its exponent's form.

These functions appear frequently in nature and economics due to their modeling of rapid growth and decay processes.

Asymptotes

Asymptotes are lines that the graph of a function approaches but never touches. They provide information about the behavior of functions at extreme values or areas of undefined behavior.

• For \( f(x) = e^{-1/x^2} \), the horizontal asymptote as \( x \to \pm \infty \) is at \( y = 1 \), reflecting that the function levels off.
• Additionally, as \( x \to 0 \), the function approaches the x-axis, indicating that the x-axis itself serves as an asymptote.

Recognizing asymptotes helps to visualize and understand the constraints and behaviors of a function.

Concavity and Inflection Points

Concavity refers to the curvature direction of a graph, while inflection points are where the graph changes its concavity. These concepts are crucial for providing a deeper understanding of the function's behavior.

• To analyze concavity, consider the second derivative. For \( f(x) = e^{-1/x^2} \), its smooth nature implies no sudden changes in concavity.
• No real inflection points exist in this function due to the uniform concave shape derived from the exponential form.

Understanding concavity and inflection points aids in identifying potential maxima, minima, and overall shape of a graph.