Problem 27 - Calculus: Graphical, Numerical, Algebraic

Question

In Exercises \(19 - 28 ,\) determine the limit graphically. Confirm algebraically. $$\lim _ { x \rightarrow 0 } \frac { \sin ^ { 2 } x } { x }$$

Step-by-Step Solution

Answer

The limit as \(x\) approaches 0 of \( \frac { \sin ^ { 2 } x } { x }\) is 0.

Step 1: Simplify the function

Rewrite the function \( \frac { \sin ^ { 2 } x } { x }\) as \( \sin x \cdot \frac { \sin x } { x }\).

Step 2: Apply the limit

As x approaches 0 for the fraction \(\frac { \sin x } { x }\), the result is 1. So the limit of our function as \(x\) approaches 0 is \(1 \cdot \lim_ { x \rightarrow 0 } \sin x\).

Step 3: Evaluate the limit

The limit of \(\sin x\) as \(x\) approaches 0 is 0. Therefore, the result of \(1 \cdot \lim_{x \rightarrow 0} \sin x = 1 \cdot 0 = 0\).

Key Concepts

Graphical analysis of limits

When analyzing limits graphically, we use the visual representation of a function to determine its behavior as it approaches a certain point. For the example \( \lim _ { x \rightarrow 0 } \frac { \sin ^ { 2 } x } { x } \), we need to look at the graph of \( y = \frac { \sin ^ { 2 } x } { x } \).
A good step is to plot the function around the critical point, which is \( x = 0 \) in this case. By observing the plot, if the function seamlessly approaches a specific value as \( x \) nears 0, we can visually approximate the limit.
For a deeper understanding, observe the upward and downward trends close to zero, and take note of any oscillatory behavior of the function. Such visual clues provide insight into the continuity and smoothness of a function at the point of interest. If the graph shows the function tapering towards a set value, that's typically your limit.

Algebraic limit confirmation

Algebraic techniques involve manipulating the expression to evaluate the limit. For \( \lim_{x \rightarrow 0} \frac { \sin ^ { 2 } x } { x } \), we simplify it first. We rewrite this as \( \sin x \cdot \frac { \sin x } { x } \).
One key algebraic strategy here is recognizing that \( \lim_{x \rightarrow 0} \frac { \sin x } { x } = 1 \). By using this fact, along with simplifying the function into the form \( \sin x \cdot 1 \), we are closer to evaluating the limit.
So, by algebraic substitution, the complex expression is reduced to a simpler form, allowing us to directly apply known limits, leading to the final simplified solution. Understanding algebraic simplification is crucial as it transforms difficult limits into manageable calculations.

Trigonometric limits

Trigonometric limits often require specific trigonometric identities or theorems for evaluation. In this exercise, recognizing the identity \( \lim_{x \rightarrow 0} \frac { \sin x } { x } = 1 \) is fundamental.
This particular limit is a standard result in calculus, often used in problems involving sine functions and their limits. By understanding and memorizing this identity, students can solve a wide range of trigonometric limit problems more efficiently.
These principles can be applied in various scenarios, often allowing the transformation of seemingly complex trigonometric expressions into much simpler ones when evaluating limits. Mastery of such trigonometric identities unlocks solutions to many calculus problems.

Limits approaching zero

When we talk about limits as a variable approaches zero, we're interested in how a function behaves near that point. Such limits can reveal insights into a function's behavior at a crucial converging point, revealing continuity and potential singularities.
In the expression \( \lim _ { x \rightarrow 0 } \frac { \sin ^ { 2 } x } { x } \), we need to carefully evaluate what happens as \( x \) gets infinitesimally close to zero.
Substituting small values of \( x \) can give us potential behavior, but rigorous analysis using algebraic and trigonometric techniques confirms the exact value. For most rational or trigonometric expressions, standard results and calculus rules, like L'Hôpital's Rule or the Squeeze Theorem, might also be applicable to find these limits accurately.