Problem 63 - Calculus A New Horizon

Question

Let \(f(x)=\left(x^{3}+1\right) / x .\) Show that the graph of \(y=f(x)\) approaches the curve \(y=x^{2}\) "asymptotically" in the sense that \(\lim _{x \rightarrow+\infty}\left[f(x)-x^{2}\right]=0 \quad\) and \(\quad \lim _{x \rightarrow-\infty}\left[f(x)-x^{2}\right]=0\) Sketch the graph of \(y=f(x)\) showing this asymptotic behavior.

Step-by-Step Solution

Answer

The graph of \(y=f(x)\) asymptotically approaches \(y=x^2\) as \(x\to\pm\infty\) since the limits of \\(f(x)-x^2\\) are 0.

Step 1: Simplify the Function

The function given is \(f(x)=\frac{x^3+1}{x}\). We can simplify this to \(f(x)=x^2+\frac{1}{x}\). This simplification helps us evaluate the limits more easily by isolating the \(x^2\) term, which we suspect the function approaches asymptotically.

Step 2: Evaluate the Limit as x Approaches Infinity

We need to find \(\lim_{x\rightarrow+\infty}\left[f(x)-x^2\right]\). Substitute the simplified form of \(f(x)\) to get \[\lim_{x\rightarrow+\infty}\left(x^2+\frac{1}{x}-x^2\right)=\lim_{x\rightarrow+\infty}\frac{1}{x}\]. As \(x\) approaches infinity, \(\frac{1}{x}\) approaches 0. Thus, the limit is \(0\).

Step 3: Evaluate the Limit as x Approaches Negative Infinity

Similarly, evaluate \(\lim_{x\rightarrow-\infty}\left[f(x)-x^2\right]\). Using the same expression \[\lim_{x\rightarrow-\infty}\left(x^2+\frac{1}{x}-x^2\right)=\lim_{x\rightarrow-\infty}\frac{1}{x}\]. As \(x\) approaches negative infinity, \(\frac{1}{x}\) also approaches 0. Thus, this limit is also \(0\).

Step 4: Interpret the Limits

Since both limits approach 0, this indicates that \(f(x)=x^2+\frac{1}{x}\) asymptotically approaches the curve \(y=x^2\) as \(x\) tends towards positive or negative infinity. This means for large magnitude \(x\), \(f(x)\) is very close to \(x^2\).

Step 5: Sketch the Graph Showing Asymptotic Behavior

Draw the graph of \(f(x)=\frac{x^3+1}{x}=x^2+\frac{1}{x}\) using a coordinate system. The curve \(y=x^2\) acts as a horizontal asymptote as \(x\rightarrow \pm\infty\). The graph of \(f(x)\) should be closely approaching \(y=x^2\) as \(x\) moves towards \(+\infty\) or \(-\infty\). Near the origin, the behavior of \(f(x)\) and \(x^2\) diverges slightly due to the \(\frac{1}{x}\) component, which dominates for smaller \(x\) values.

Key Concepts

Limit Evaluation

When working with limits, especially as they approach infinity, we are interested in finding how a function behaves or what value it nears. Let's consider the function \(f(x) = \frac{x^3 + 1}{x}\). By simplifying it to \(f(x) = x^2 + \frac{1}{x}\), the function becomes more manageable, because \(x^2\) can be treated separately from the diminishing term \(\frac{1}{x}\).
For limit evaluation as \(x\) approaches infinity, we compute \(\lim_{x\rightarrow +\infty}(f(x) - x^2)\). Substituting our simplified version of \(f(x)\), this becomes \(\lim_{x\rightarrow +\infty}(x^2 + \frac{1}{x} - x^2) = \lim_{x\rightarrow +\infty}\frac{1}{x}\). As \(x\) gets larger, \(\frac{1}{x}\) gets closer to 0. Thus, the limit evaluates to 0.
Similarly, for \(x\) approaching negative infinity, we have \(\lim_{x\rightarrow -\infty}(f(x) - x^2)\). Again, substituting gives \(\lim_{x\rightarrow -\infty}(x^2 + \frac{1}{x} - x^2) = \lim_{x\rightarrow -\infty}\frac{1}{x}\). Here too, as \(x\) heads to negative infinity, \(\frac{1}{x}\) approaches 0, so the limit is 0. These calculations demonstrate that \(f(x)\) aligns closely with \(y = x^2\) as \(x\) moves outward in either direction, confirming the asymptotic nature.

Graph Sketching

Graph sketching is a valuable skill in understanding the behavior of functions. It visually represents the relationship between the function's domains and ranges. Let's break down how to accurately depict \(f(x) = \frac{x^3 + 1}{x}\), or the equivalent \(f(x) = x^2 + \frac{1}{x}\).
Firstly, plot the curve \(y = x^2\). This gives us a baseline guide as this parabolic curve acts as a kind of asymptotic guide for \(f(x)\). Next, for the same \(x\) values, plot \(f(x)\). Pay special attention to the behavior around the origin, because there's a divergence as \(f(x)\) shifts more noticeably from \(y = x^2\) at small \(x\) values, due to the impact of the \(\frac{1}{x}\) term.
As \(x\) increases towards infinity or decreases towards negative infinity, you will notice \(f(x)\) gets closer and closer to \(y = x^2\), illustrating the concept of horizontal asymptotes as \(x^2\). The graph should clearly depict how the function adjusts for small \(x\) and how it stabilizes, ensuring the proper asymptotic behavior is observed as \(x\) goes to either direction without bound.

Calculus Problem Solving

Solving calculus problems typically involves breaking down complex problems into simpler steps. Let's consider the example of asymptotic behavior using limits and simplification.
Initially, simplifying \(f(x) = \frac{x^3 + 1}{x}\) to \(f(x) = x^2 + \frac{1}{x}\) is crucial. This is because it separates the core element \(x^2\) from the minor perturbation \(\frac{1}{x}\). Such simplifications make evaluating limits more straightforward and reveal the true behavior of functions at infinity.
Understanding limits in this context means identifying terms that become negligible as \(x\) grows large or negative. This approach directly assists in showing how functions behave with respect to other simpler curves, like parabolas, very far out on the graph.
To solve calculus problems effectively, focus on:

• Identifying dominant terms in the function.
• Recognizing the effects of small perturbations, like \(\frac{1}{x}\).
• Breaking the problem into evaluations for positive and negative infinities separately.
• Using graphical insight to confirm analytical results.
• Each step contributes to a comprehensive understanding and solution of the problem.

By following these strategies, one can find solutions to calculus problems while gaining insight into their deeper behaviors, paving the way for understanding a wide array of phenomena in mathematics.