Question

Show that \(\lim _{x=\infty} \frac{x^{2}+1}{3 x^{2}+2 x-1}=\frac{1}{3}\).

Short answer

Answer: The limit of the function as \(x\) approaches infinity is \(\frac{1}{3}\).

Step-by-step solution

Analyze the given expression

We are given the expression \(\frac{x^2+1}{3x^2+2x-1}\) and we need to find its limit as \(x\) approaches infinity.

Simplify the expression

To simplify the expression, divide both the numerator and the denominator by the highest power of \(x\) present in the expression, which is \(x^2\). This will result in: \(\frac{\frac{x^2}{x^2}+\frac{1}{x^2}}{\frac{3x^2}{x^2}+\frac{2x}{x^2}-\frac{1}{x^2}}\). Now, simplify each term: \(\frac{1+\frac{1}{x^2}}{3+\frac{2}{x}-\frac{1}{x^2}}\).

Apply the properties of limits

Now, we want to find the limit as \(x\) approaches infinity: \(\lim _{x=\infty} \frac{1+\frac{1}{x^2}}{3+\frac{2}{x}-\frac{1}{x^2}}\). Since the limit of a sum or difference is equal to the sum or difference of the limits, we can find the individual limits: \\ \(\lim_{x=\infty} 1 = 1\), \\ \(\lim_{x=\infty} \frac{1}{x^2} = 0\), \\ \(\lim_{x=\infty} 3 = 3\), \\ \(\lim_{x=\infty} \frac{2}{x} = 0\), \\ \(\lim_{x=\infty} \frac{1}{x^2} = 0\). \\ Now, substitute these values back into the expression: \(\frac{1+0}{3+0-0} = \frac{1}{3}\).

Conclusion

As a result, we have shown that \(\lim _{x=\infty} \frac{x^{2}+1}{3 x^{2}+2 x-1}=\frac{1}{3}\).

Key concepts

Rational Functions

A rational function is a fraction where both the numerator and the denominator are polynomials. For example, the expression \( \frac{x^2 + 1}{3x^2 + 2x - 1} \) is a typical rational function, as both the top and bottom are polynomial expressions. Rational functions are common in calculus because they easily illustrate the concept of limits and behavior as values grow very large or very small.

The beauty of rational functions lies in their predictability and regular patterns, especially at extreme values. In our case, both the numerator \(x^2 + 1\) and the denominator \(3x^2 + 2x - 1\) have quadratic terms as their highest degree terms. Understanding the degree of these terms helps predict how the function behaves at the limits of infinity.

• Degree of the numerator: the highest power of \(x\), here \(x^2\).
• Degree of the denominator: same as the numerator, which is \(x^2\).

When both parts of a rational function have the same degree, like in our expression, simple techniques can help find the limit of the function at infinity. This simplifies calculus problems, helping us determine if the function levels off to a fixed value, grows indefinitely, or approaches zero.

Limits at Infinity

The limit of a function as it approaches infinity is investigating how the function behaves as its input grows very large. For rational functions like \( \frac{x^2 + 1}{3x^2 + 2x - 1} \), determining the limit at infinity helps us predict the function's behavior in extreme cases.

To tackle limits at infinity, it's common to compare the degrees of the numerator and the denominator in a rational function. The three main possibilities are:

• If the numerator has a higher degree than the denominator, the limit at infinity is often infinite, or the function increases or decreases without bound.
• If the numerator's degree is less than the denominator's, the limit at infinity typically results in zero.
• If both are of equal degree, like in our problem, the limit is determined by dividing the leading coefficients of both the numerator and the denominator. In the expression \(\frac{x^2 + 1}{3x^2 + 2x - 1}\), this means dividing the lead coefficient \(1\) by \(3\), giving \(\frac{1}{3}\) as the limit.

These guiding principles allow for quick evaluation of behavior in the great beyond of large \(x\) values. Our focus is to get a simplified version of the expression, which is more understandable.

Simplifying Expressions

Simplifying expressions involves reducing them to their simplest form to make mathematical operations like finding limits easier. In our context of finding a limit at infinity, simplifying the expression \( \frac{x^2 + 1}{3x^2 + 2x - 1} \) is crucial.

To begin simplifying, focus on the highest power of \(x\) in both the numerator and the denominator. Here, it's \(x^2\). By dividing every term by \(x^2\), the expression becomes:
\[ \frac{\frac{x^2}{x^2} + \frac{1}{x^2}}{\frac{3x^2}{x^2} + \frac{2x}{x^2} - \frac{1}{x^2}} \rightarrow \frac{1 + \frac{1}{x^2}}{3 + \frac{2}{x} - \frac{1}{x^2}} \]
Using this form, we can identify terms that vanish as \(x\to\infty\). For instance, terms like \( \frac{1}{x^2} \) and \( \frac{2}{x} \) shrink to zero because \(x\) in the denominator means they get smaller and smaller.

Finally, substituting back these simplified limits to calculate the limit at infinity gives us \(\frac{1+0}{3+0-0} = \frac{1}{3}\). Simplifying like this not only aids in solving the problem but also enhances fundamental understanding of behavior in calculus. Simplification demystifies complex expressions, reducing them to a form that clearly shows how each component influences the overall result.