Problem 54 - Calculus: Early Transcendentals

Question

(a) Graph the function $$ f(x) = \frac{\sqrt{2x^2 + 1}}{3x - 5} $$ How many horizontal and vertical asymptotes do you observe? Use the graph to estimate the values of the limits $$ \lim_{x \to \infty} \frac{\sqrt{2x^2 + 1}}{3x - 5} \hspace{5mm} \text{and} \hspace{5mm} \lim_{x \to -\infty} \frac{\sqrt{2x^2 + 1}}{3x - 5} $$ (b) By calculating values of \( f(x) \), give numerical estimates of the limits in part (a). (c) Calculate the exact values of the limits in part (a). Did you get the same value or different values for these two limits? [In view of your answer to part (a), you might have to check your calculation for the second limit.]

Step-by-Step Solution

Answer

One vertical asymptote at \(x = \frac{5}{3}\); horizontal limits are \(\frac{\sqrt{2}}{3}\) and \(-\frac{\sqrt{2}}{3}\). Limits agree with estimates.

Step 1: Identify potential vertical asymptotes

To find the vertical asymptotes, set the denominator of the function equal to zero, as the function is undefined where the denominator is zero. Solve \(3x - 5 = 0\), yielding \(x = \frac{5}{3}\). Thus, there is a vertical asymptote at \(x = \frac{5}{3}\).

Step 2: Determine horizontal asymptotes

We calculate the horizontal asymptotes by evaluating the limits at infinity. Divide each term in the numerator by \(x\): \[\lim_{x \to \infty} \frac{\sqrt{2x^2 + 1}}{3x - 5} = \lim_{x \to \infty} \frac{\sqrt{x^2(2 + \frac{1}{x^2})}}{3x}\]. Simplifying gives \(\frac{\sqrt{2}}{3}\). A similar step shows \(\lim_{x \to -\infty} \frac{\sqrt{2}}{-3}\).

Step 3: Graphing the function

Create a graph of \(f(x) = \frac{\sqrt{2x^2 + 1}}{3x - 5}\). Observe the behavior as \(x\) approaches the vertical asymptote \(x = \frac{5}{3}\) and watch the function's trend toward the horizontal asymptotes \(\frac{\sqrt{2}}{3}\) as \(x \to \infty\) and \(-\frac{\sqrt{2}}{3}\) as \(x \to -\infty\).

Step 4: Estimate limits numerically

Select values such as \(x = 100, 1000\) and \(-100, -1000\) to approximate \(f(x)\) as \(x\) approaches \(\infty\) and \(-\infty\). Observing numerical values, the estimates closely align with the horizontal asymptotes, supporting the calculated limits in Step 2.

Step 5: Verify exact limit calculations

Confirm the limit calculations: Using the simplified expression from Step 2, the limits are: \(\lim_{x \to \infty} \frac{\sqrt{x^2(2 + \frac{1}{x^2})}}{3x} = \frac{\sqrt{2}}{3}\) and \(\lim_{x \to -\infty} \frac{\sqrt{x^2(2 + \frac{1}{x^2})}}{3x} = -\frac{\sqrt{2}}{3}\), thus agreeing with the graph and numerical estimates.

Key Concepts

Horizontal Asymptotes

When analyzing horizontal asymptotes, we aim to find the values that a function approaches as the variable moves towards positive or negative infinity. In simpler terms, it's the horizon line where the function seems to settle. For the function \( f(x) = \frac{\sqrt{2x^2 + 1}}{3x - 5} \), the degree of the highest power term in the numerator is \(x^2\) inside the square root. The denominator has the degree \(x\).

To identify horizontal asymptotes, we compared the degrees of the variable in the numerator and denominator. Simplifying \( \lim_{x \to \infty} \frac{\sqrt{2x^2 + 1}}{3x - 5} \) by dividing through by \(x\) gives \( \lim_{x \to \infty} \frac{\sqrt{x^2(2 + \frac{1}{x^2})}}{3x}\), which results in \(\frac{\sqrt{2}}{3}\). Similarly, for \(\lim_{x \to -\infty} \), the asymptote becomes \(-\frac{\sqrt{2}}{3}\). Therefore, this function has two horizontal asymptotes.

Horizontal asymptotes help us understand the behavior of functions as they extend far beyond any visible limits on a graph.

Vertical Asymptotes

Vertical asymptotes represent the vertical lines where the function values rise or fall indefinitely. These occur at points where the function is undefined, typically where the denominator equals zero in a rational expression.

For \( f(x) = \frac{\sqrt{2x^2 + 1}}{3x - 5} \), set the denominator to zero to find vertical asymptotes. Solving \(3x - 5 = 0\), we find the asymptote at \(x = \frac{5}{3}\).

On a graph, you'll notice that as \(x\) approaches this point from either side, the function values diverge to \(\pm \infty\), creating a clear break or gap in the graph's continuity. Vertical asymptotes indicate a boundary of sorts, beyond which the function does not exist.

Limit Calculation

Limit calculation is a fundamental part of understanding the behavior of functions at specific points, especially across asymptotes. Limits allow us to predict trends of a function without explicitly reaching or exceeding those boundaries.

For the given function, we calculated the limits at \( \infty \) and \(-\infty\). Detailed steps involved dividing by the highest power of \(x\) present. This simplification helps reveal the core trend and leads to the recognized horizontal asymptotes. Through this process, \( \lim_{x \to \infty} \frac{\sqrt{2x^2 + 1}}{3x - 5} \) becomes simply \(\frac{\sqrt{2}}{3}\), accurate to many numerical estimations. The opposite extreme, \( \lim_{x \to - \infty} \), results in \(-\frac{\sqrt{2}}{3}\).

Mastering limits is crucial for unraveling intricate function behaviors, making it an anchor concept in calculus.

Infinite Limits

Infinite limits occur in scenarios where function values grow indefinitely as \(x\) approaches a set point, often at a vertical asymptote. This contrasts with horizontal limits, where functions settle towards a constant value.

Investigating \( f(x) = \frac{\sqrt{2x^2 + 1}}{3x - 5} \), we observe an infinite limit approaching the vertical asymptote at \(x = \frac{5}{3}\). Approaching from both the left and right, the function \(f(x)\) tends toward \(\infty\) in one direction and \(-\infty\) in the other, clearly denoting the vertical asymptote.

Infinite limits help determine behaviors that contain discontinuities or rapid increases, crucial for predicting a function’s divergence or convergence in practical scenarios.