Question

Determine the following limits. $$\lim _{x \rightarrow-\infty} \frac{\sqrt{16 x^{2}+x}}{x}$$

Short answer

Answer: The limit is 0.

Step-by-step solution

Factor out x from the numerator

We can start by factoring out x from the numerator of the fraction inside the limit. To do this, note that \(16x^2+x = x(16x+1)\). So, we can rewrite the expression inside the limit as follows: $$\lim _{x \rightarrow -\infty} \frac{\sqrt{x(16x+1)}}{x}$$

Factor x out of the square root

Next, we want to factor x out of the square root so that we can simplify the expression further. To do this, recall that \(\sqrt{ab}=\sqrt{a}\sqrt{b}\). Therefore, we have: $$\lim _{x \rightarrow -\infty} \frac{\sqrt{x}\sqrt{16x+1}}{x}$$

Cancel the x term in the numerator and denominator

Now, we can cancel out the x term in the numerator and denominator as follows: $$\lim _{x \rightarrow -\infty} \frac{\sqrt{x}\sqrt{16x+1}}{x} = \lim _{x \rightarrow -\infty} \frac{1\cdot\sqrt{16x+1}}{x} \cdot \frac{\sqrt{x}}{\sqrt{x}} = \lim _{x \rightarrow -\infty} \frac{\sqrt{16x+1}}{\sqrt{x}}$$

Rationalize the denominator

In order to find the limit of this expression, we need to rationalize the denominator. To do this, we will multiply both the numerator and the denominator by \(\sqrt{x}\): $$\lim _{x \rightarrow -\infty} \frac{\sqrt{16x+1}}{\sqrt{x}} \cdot \frac{\sqrt{x}}{\sqrt{x}} = \lim _{x \rightarrow -\infty} \frac{\sqrt{16x^2+x^{2}}}{x^2} = \lim _{x \rightarrow -\infty} \frac{\sqrt{17x^2}}{x^2}$$

Simplify the expression and find the limit

Now, we can simplify the expression inside the limit by dividing \(\sqrt{17x^2}\) by \(x^2\). $$\lim _{x \rightarrow -\infty} \frac{\sqrt{17x^2}}{x^2} = \lim _{x \rightarrow -\infty} \frac{\sqrt{17}\cdot\sqrt{x^2}}{x^2} = \lim _{x \rightarrow -\infty} \frac{\sqrt{17}\cdot|x|}{x^2}$$ As \(x\) approaches \(-\infty\), \(|x|\) will approach \(-\infty\cdot -1 = \infty\) and \(x^2\) will approach \(\infty\). Therefore, the limit is: $$\lim _{x \rightarrow -\infty} \frac{\sqrt{17}\cdot|x|}{x^2} = \frac{\sqrt{17}\cdot\infty}{\infty^2} = \frac{\infty}{\infty} = 0$$ So, the limit of the given expression as \(x\) approaches negative infinity is 0.

Key concepts

Infinite Limits

When we talk about infinite limits, we are interested in what happens to a function as the variable approaches infinity or negative infinity. In the exercise provided, we examine the behavior of the function \(\frac{\sqrt{16x^2+x}}{x}\) as \(x\) goes to \(-\infty\). This means we're checking how the function's output changes when \(x\) becomes very large negatively. Think of infinite limits like watching a car zoom into the horizon on a long road.

To understand these limits:

• Identify how the dominant terms of the expressions behave as \(x\) becomes very large or very small.
• Consider if the function grows without bounds, approaches zero, or stabilizes to a finite value.

In this case, simplifying and analyzing dominant terms showed that as \(x\) approaches negative infinity, the fraction within the square root becomes negligible compared to the term with \(x^2\), leading to the outcome of 0.

Rationalization Techniques

Rationalization is a technique used to eliminate radicals (like square roots) in the denominator of an expression. This makes the limits easier to evaluate and simplify. In our exercise, the technique was applied to the expression \(\frac{\sqrt{16x+1}}{\sqrt{x}}\) by multiplying both the numerator and denominator by \(\sqrt{x}\) to clear the radical.

Why rationalize? Because having radicals in the denominator can make evaluating limits tricky. Here's what you do:

• Multiply the numerator and denominator by the conjugate of the denominator's radical.
• This transforms the denominator into a rational expression, simplifying limit calculation.

In the example, multiplying through by \(\sqrt{x}\) changed the expression into \(\frac{\sqrt{17x^2}}{x^2}\), allowing for further simplification and analysis.

Simplifying Expressions

Simplifying expressions is crucial to make limits more understandable and easier to solve. In the given problem, it's imperative to break down and rewrite complex terms for simplification. This exercise involved steps like factoring out \(x\), acknowledging the power of dominant terms, and canceling unnecessary components.

Here's a general approach:

• Identify common factors or terms to simplify the expression.
• Use algebraic identities, such as \(\sqrt{a}\sqrt{b} = \sqrt{ab}\).
• Factor out terms to magnify the influence of dominant terms.

By recognizing that \(16x^2+x\) simplifies to \(x(16x+1)\), we eliminated complex calculations, reducing the expression to something more manageable, ultimately showing that the simplified limit result is 0.

Limit Laws

Limit laws are rules that simplify evaluating limits and make them more predictable to solve. These laws cover operations like addition, subtraction, multiplication, division, and finding powers of limits.

In the provided exercise, several key limit laws were implicitly used:

• Quotient Law: Used when dividing functions. If the limit of the denominator is nonzero, the limit of the quotient is the limit of the quotient of the limits.
• Multiplication by a Constant: Constants like \(\sqrt{17}\) factor out, maintaining the structure of the limit as \(x\) changes.
• Power Law: Relevant when dealing with powers, as observed in simplifying \(\sqrt{17x^2}\).

In our case, knowing these laws helped in systematically reducing and evaluating the limit, streamlining the path to confirming that the value is indeed 0 as \(x\) approaches \(-\infty\).