Problem 14 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \lim _{x \rightarrow 0^{-}} \frac{3 \sin x}{\sqrt{-x}} $$

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Understand the Limit Notation

The goal is to find the limit of the function \( \frac{3 \sin x}{\sqrt{-x}} \) as \( x \) approaches 0 from the left, indicated by \( x \to 0^- \). This means \( x \) is negative, and approaching zero.

Step 2: Analyze the Expression

For \( x \to 0^- \), \( \sin x \) is approximately \( x \) because \( \sin x \approx x \) when \( x \) is near 0. Substitute \( \sin x \) with \( x \) for simplicity.

Step 3: Simplify the Function for the Limit

By substituting \( \sin x \approx x \), the expression becomes \( \frac{3x}{\sqrt{-x}} \). Simplify this to \( \frac{3x}{i\sqrt{x}} \), since \( \sqrt{-x} = i\sqrt{x} \), where \( i \) is the imaginary unit.

Step 4: Further Simplification

We can simplify \( \frac{3x}{i\sqrt{x}} \) by rewriting it as \( \frac{3}{i} \cdot \sqrt{x} \). Here, the division by \( i \) is retained, and \( x^{-1/2} \) is simplified.

Step 5: Evaluate the Limit

As \( x \to 0^- \), \( \sqrt{x} \to 0 \). Thus, the expression \( \frac{3}{i} \cdot \sqrt{x} \) also tends to \( 0 \). The limit of \( \frac{3 \sin x}{\sqrt{-x}} \) as \( x \to 0^- \) is 0.

Key Concepts

Left-Hand Limit

When talking about limits in calculus, the idea is to find out what value a function approaches as the input gets closer to a certain point. A **left-hand limit** specifically looks at the values as they approach from the left side. For our problem, we have \( x \to 0^- \), meaning \( x \) comes from slightly negative values. This is crucial because it lets us only consider the values just to the left of zero. From this perspective, we analyze how the function behaves as these inputs get closer to zero.It’s important because functions can behave differently on the left and right sides of a point. The left-hand limit helps us understand any unique characteristics or behaviors of the function, particularly important in cases with potential discontinuities or as in this exercise, when dealing with square roots of negative values.

Simplifying Trigonometric Expressions

Trigonometric functions like sine and cosine are often simplified when we are finding limits, especially as the variable approaches zero. Here, we use a known limit property: when \( x \) is close to zero, \( \sin x \approx x \). This approximation allows us to replace \( \sin x \) with \( x \), simplifying our calculations significantly.Here's why this simplification is useful:

• Trigonometric functions can be complex, so using approximations streamlines calculations.
• For functions approaching zero, approximating \( \sin x \) to \( x \) is mathematically sound and makes softer the exploration of behavior near zero.

This simplification transforms our initial function, easing the work involved in solving the limit problem as seen in the solution provided. It forms the backbone of simplifying expressions and successfully evaluating limits.

Imaginary Numbers

Imaginary numbers come into play when working with square roots of negative numbers. In this exercise, while simplifying \( \sqrt{-x} \), we use imaginary numbers. By definition, the imaginary unit \( i \) is defined as \( i=\sqrt{-1} \). Therefore, \( \sqrt{-x} = i\sqrt{x} \).Here's why understanding imaginary numbers is essential:

• They allow us to extend calculus and algebra into the complex plane, solving problems with negative square roots.
• Imaginary units help simplify expressions that may not be solvable in the real number system alone.

In our problem, acknowledging the imaginary component lets us reframe the equation correctly, keep calculations precise, and helps us accurately determine the limit's behavior. Imaginary numbers convert seemingly unsolvable expressions into manageable problems, a crucial step in complex analysis.