Question

Evaluate \(\lim _{x \rightarrow \infty} f(x)\) and \(\lim _{x \rightarrow-\infty} f(x)\) for the following functions. Then give the horizontal asymptote(s) of \(f\) (if any). $$f(x)=4 x(3 x-\sqrt{9 x^{2}+1})$$

Short answer

Answer: As $$x$$ approaches positive and negative infinity, the function $$f(x)$$ approaches infinity, and there are no horizontal asymptotes.

Step-by-step solution

Analyze the function

The function $$f(x)$$ is a product of two terms: $$4x$$ and $$(3x - \sqrt{9x^2 + 1})$$. We can find the limits for each term separately and then use the limit product rule to find the limit of the overall function.

Find the limit of the first term

The first term is $$4x$$. As $$x$$ approaches positive infinity, $$4x$$ also approaches positive infinity, and as $$x$$ approaches negative infinity, $$4x$$ approaches negative infinity.

Simplify the second term

Rewrite the second term by factoring out an $$x$$: $$(3x - \sqrt{9x^2 + 1}) = x(3 - \frac{\sqrt{9x^2 + 1}}{x})$$

Find the limit of the simplified second term

We can now analyze the limit of the simplified second term as $$x$$ approaches infinity: As $$x$$ approaches positive or negative infinity: $$\lim_{x \to \pm\infty} (3 - \frac{\sqrt{9x^2 + 1}}{x})$$ Divide the terms inside the square root by $$x^2$$: $$\lim_{x \to \pm\infty} (3 - \frac{\sqrt{9 + \frac{1}{x^2}}}{x})$$ Now, as $$x$$ approaches positive or negative infinity, the term $$\frac{1}{x^2}$$ approaches 0: $$\lim_{x \to \pm\infty} (3 - \frac{\sqrt{9}}{x}) = \lim_{x \to \pm\infty} (3 - \frac{3}{x})$$ As $$x$$ approaches positive or negative infinity, the term $$\frac{3}{x}$$ approaches 0: $$\lim_{x \to \pm\infty} (3 - \frac{3}{x}) = 3$$

Find the limits of the overall function

Using the limit product rule, we can find the limits of the overall function: $$\lim_{x \to \infty} f(x) = (\lim_{x \to \infty} 4x)(\lim_{x \to \infty} x(3 - \frac{3}{x}))$$ Since both terms approach infinity, the limit as $$x$$ approaches positive infinity is infinity: $$\lim_{x \to \infty} f(x) = \infty$$ Similarly, for the limit as $$x$$ approaches negative infinity: $$\lim_{x \to -\infty} f(x) = (\lim_{x \to -\infty} 4x)(\lim_{x \to -\infty} x(3 - \frac{3}{x}))$$ Since both terms approach negative infinity, the limit as $$x$$ approaches negative infinity is positive infinity: $$\lim_{x \to -\infty} f(x) = \infty$$

Horizontal asymptotes

Since the limits as $$x$$ approaches positive and negative infinity are both $$\infty$$, there are no horizontal asymptotes for the function $$f(x)$$. In conclusion, the limits of the function $$f(x)$$ are: $$\lim_{x \to \infty} f(x) = \infty$$ $$\lim_{x \to -\infty} f(x) = \infty$$ And the function has no horizontal asymptotes.

Key concepts

Horizontal Asymptotes

Horizontal asymptotes help us understand the behavior of a function as the input values become extremely large or small. Specifically, these are horizontal lines that the graph of a function approaches as the input goes to infinity or negative infinity.

For a function like \(f(x)\), if \(\lim_{x \to \infty} f(x) = L\) and \(\lim_{x \to -\infty} f(x) = L\), then \(y = L\) is a horizontal asymptote of the function. This indicates that as \(x\) becomes very large or very small, \(f(x)\) gets closer and closer to \(L\), but doesn't necessarily touch or cross this line.

In our original problem, both limits as \(x\) approaches infinity and negative infinity resulted in infinity. This means the function grows indefinitely large in both directions and therefore does not have horizontal asymptotes.

• A horizontal asymptote can be seen as a long-term destination that a function moves towards but perhaps never reaches.
• Functions exhibiting different behaviors at opposite extremes may have different limits and thus different horizontal asymptotes or none at all.

Infinity

Infinity is a concept that describes something without any bound or limit. In mathematics, infinity is not a number but a representation of an unbounded quantity. It plays a significant role when discussing limits and asymptotes.

In the context of the original exercise, when examining the behavior of \(f(x)\) at extreme \(x\) values, we found both \(\lim_{x \to \infty} f(x) = \infty\) and \(\lim_{x \to -\infty} f(x) = \infty\). Here is what this means:

• Positive infinity: As \(x\) becomes larger and larger (positive infinity), \(f(x)\) increases without bound.
• Negative infinity: As \(x\) becomes more and more negative (negative infinity), \(f(x)\) also increases without bound.

Understanding infinity helps us make sense of how functions behave over wide intervals and recognize what the end behavior might look like.

Limit Product Rule

The Limit Product Rule is a simple yet powerful principle in calculus. It states that the limit of a product of two functions is equal to the product of their limits—provided these limits exist. This rule can simplify complex limit evaluations in calculus.

In our example exercise, the function \(f(x) = 4x(3x - \sqrt{9x^2 + 1})\) requires applying the Limit Product Rule to determine its behavior as \(x\) moves towards positive and negative infinity.

• First term: \(4x\) simplifies easily as \(x\) goes to infinity, contributing simply as \(\pm \infty\).
• Second term: Needs simplification and further evaluation to see how it behaves, as done in the solution steps to break it into manageable limits.

By separating the function into manageable parts, applying the product rule directly simplifies the evaluation.

Ultimately, recognizing where and how to apply the Limit Product Rule can reveal critical insights about the function's behavior over infinite intervals—a crucial skill in mastering calculus concepts.