Problem 28 - Calculus of a Single Variable

Question

Finding a Limit In Exercises \(19-38,\) find the limit. $$ \lim _{x \rightarrow-\infty} \frac{x}{\sqrt{x^{2}+1}} $$

Step-by-Step Solution

Answer

The limit \( \lim _{x \rightarrow-\infty} \frac{x}{\sqrt{x^{2}+1}} = 1.

Step 1: Set Up the Limit

The function whose limit we need to find is \( \frac{x}{\sqrt{x^{2}+1}} \) and \( x \) is approaching negative infinity. Hence, the limit to find is \( \lim _{x \rightarrow -\infty} \frac{x}{\sqrt{x^{2}+1}} \).

Step 2: Multiplication by the Conjugate

To simplify this limit, multiply the numerator and denominator by the conjugate of the denominator \( \sqrt{x^{2}+1} \), which is \( -\sqrt{x^{2}+1} \). This gives us \( \lim _{x \rightarrow-\infty} \frac{x(-\sqrt{x^{2}+1})}{(\sqrt{x^{2}+1})(-\sqrt{x^{2}+1})} \).

Step 3: Simplifying the Expression

We simplify the expression and it turns out to be \( \lim _{x \rightarrow-\infty} \frac{-x\sqrt{x^{2}+1}}{x^{2}-1} \).

Step 4: Applying Lim Property

The limit of the quotient equals the quotient of the limits if the limit of the denominator isn't zero, thus splitting the limit up gives \( \frac{\lim _{x \rightarrow-\infty} (-x\sqrt{x^{2}+1})}{\lim _{x \rightarrow-\infty} (x^{2}-1)} \).

Step 5: Detailed Calculation for Both Limits

Calculate the two limits independently. For the numerator, if we input \( -\infty \) we get \( -(-\infty) = \infty \), for the denominator we have \( (-\infty^{2}) - 1 = \infty - 1 = \infty \). Therefore, both limits are \( \infty \) and the limit of the fraction is the form \( \frac{\infty}{\infty} \), an indeterminate form.

Step 6: L'Hopital's Rule

We can apply L'Hopital's rule which states, when the limit is in indeterminate form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), the limit of the fraction is equal to the limit of the derivatives of the numerator and the denominator. Therefore, the function turns out to be \( \lim _{x \rightarrow-\infty} \frac{-\sqrt{x^{2}+1} - \frac{x^{2}}{\sqrt{x^{2}+1}}}{2x} \).

Step 7: Simplifying the Function

Simplify the function to \( \lim _{x \rightarrow-\infty} \frac{-\sqrt{x^{2}+1}}{2x} \). We substitute the limit \( x \rightarrow -\infty \), which results in \( \frac{-\sqrt{(-\infty)^{2}+1}}{2(-\infty)} \) = \( \frac{-\infty}{-\infty} \) = 1.

Key Concepts

Limit at Infinity

When we discuss limits at infinity, we're examining what happens to a function as the variable approaches a very large positive or negative number. It essentially gives us an idea of the function's behavior or tendency as it reaches towards infinite values. In the context of the exercise, we are evaluating the limit of the function \( \frac{x}{\sqrt{x^2 + 1}} \) as \( x \) tends towards negative infinity.

This means we investigate the expression as \( x \) becomes a very large negative number, which helps us identify how the function behaves in extreme situations. It’s crucial to understand that a limit at infinity provides information about the end behavior of the function rather than its exact values.

Indeterminate Forms

An indeterminate form often arises within calculus problems and is a type of expression where the limits are not readily apparent. Commonly encountered indeterminate forms are \( \frac{0}{0} \) and \( \frac{\infty}{\infty} \). In this exercise, after evaluating both the numerator and the denominator separately by substituting \( x \to -\infty \), we initially arrived at the indeterminate form \( \frac{\infty}{\infty} \).

It's significant because having an indeterminate form implies that conventional limits can't be immediately determined, and thus, further simplification or specific techniques like L'Hôpital's Rule are necessary to find the true limit. Understanding this concept is pivotal in recognizing when a simple limit calculation won't suffice, requiring more advanced methods.

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool in calculus used to resolve limits where indeterminate forms such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) occur. According to this rule, if a limit turns out to be indeterminate, the limit of the original function can be determined by finding the limit of the derivatives of its numerator and denominator.

In our exercise, the form \( \frac{\infty}{\infty} \) prompted us to apply L'Hôpital's Rule. By taking the derivatives of both the numerator and the denominator and then applying the limit to this new function, we simplify the situation, allowing us to determine the limit isn't just an indeterminate form but has a value as the approach progresses.

Limit Simplification

Once you've applied techniques like L'Hôpital's Rule during complex limits involving indeterminate forms, simplification is often necessary to find the final answer. Simplifying expressions helps in breaking down a complicated mathematical problem into manageable parts.

In our case with \( \frac{x}{\sqrt{x^2 + 1}} \), simplification after application of L'Hôpital's Rule and re-evaluation led to a manageable form that allowed straightforward substitution and calculation. Recognizing opportunities for simplification during limit evaluation helps you grasp the core behavior of even the complex functions easily, leading to a clear solution of the problem.