Problem 40
Question
Challenge your understanding of limits but can be evaluated using the knowledge gained in this section. $$ \lim _{x \rightarrow 0} \frac{\sin 5 x}{8 x} $$
Step-by-Step Solution
Verified Answer
The limit is \( \frac{5}{8} \).
1Step 1: Identify the Limit Form
The given expression is \( \lim _{x \rightarrow 0} \frac{\sin 5x}{8x} \). As \( x \) approaches 0, both the numerator and denominator approach 0, leading to an indeterminate form \( \frac{0}{0} \). Therefore, we need to simplify this expression to evaluate the limit.
2Step 2: Use the Limit Property of Sine
We know from limit properties that \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \). To use this, we need the argument of sine to match the denominator. Rewrite the expression as follows: \[ \lim _{x \rightarrow 0} \frac{\sin 5x}{5x} \cdot \frac{5x}{8x} = \lim _{x \rightarrow 0} \left(\frac{\sin 5x}{5x}\right) \cdot \frac{5}{8} \]
3Step 3: Evaluate the Simplified Limit
Now, separately evaluate the limits:\( \lim_{x \to 0} \frac{\sin 5x}{5x} = 1 \) (using the standard limit property). The rest of the expression is \( \frac{5}{8} \). Therefore, the evaluated limit is:\[ \lim _{x \rightarrow 0} \frac{\sin 5x}{8x} = 1 \cdot \frac{5}{8} = \frac{5}{8} \]
Key Concepts
LimitsIndeterminate FormsTrigonometric Limits
Limits
When we talk about limits in calculus, we're essentially exploring what happens to a function as its input approaches a particular point. In the exercise we're working on, we are interested in how the expression \( \frac{\sin 5x}{8x} \) behaves as \( x \) gets closer to zero.
Limits play a crucial role in calculus as they form the foundation for defining derivatives and integrals. They help us understand continuous behavior even when the function might not be defined at certain points. In this case, as \( x \) approaches 0, we try to comprehend the behavior of the quotient \( \frac{\sin 5x}{8x} \).
Finding the limit of such expressions often involves simplification, especially when confronted with potential issues like division by zero. It's a valuable technique to foresee the function's behavior, enabling us to deduce meaningful conclusions even from indeterminate forms.
Limits play a crucial role in calculus as they form the foundation for defining derivatives and integrals. They help us understand continuous behavior even when the function might not be defined at certain points. In this case, as \( x \) approaches 0, we try to comprehend the behavior of the quotient \( \frac{\sin 5x}{8x} \).
Finding the limit of such expressions often involves simplification, especially when confronted with potential issues like division by zero. It's a valuable technique to foresee the function's behavior, enabling us to deduce meaningful conclusions even from indeterminate forms.
Indeterminate Forms
In our exercise, the expression \( \lim_{x \rightarrow 0} \frac{\sin 5x}{8x} \) leads to an indeterminate form of \( \frac{0}{0} \). These forms arise when both the numerator and denominator of a fraction approach zero as the variable approaches a certain value, leading to ambiguity.
Indeterminate forms can seem perplexing because they don't lead to straightforward mathematical conclusions. However, employing limit properties and algebraic manipulation helps unravel their mysteries. Techniques include factoring, using conjugates, or employing trigonometric identities as we did here.
In the exercise, we used the known property \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \) to simplify and resolve the indeterminate form. By adjusting the original expression to match this property, we unlocked a clear answer, moving past the initial impasse of indeterminacy. It's a perfect example of dissecting seemingly unsolvable problems using clever mathematical tools.
Indeterminate forms can seem perplexing because they don't lead to straightforward mathematical conclusions. However, employing limit properties and algebraic manipulation helps unravel their mysteries. Techniques include factoring, using conjugates, or employing trigonometric identities as we did here.
In the exercise, we used the known property \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \) to simplify and resolve the indeterminate form. By adjusting the original expression to match this property, we unlocked a clear answer, moving past the initial impasse of indeterminacy. It's a perfect example of dissecting seemingly unsolvable problems using clever mathematical tools.
Trigonometric Limits
Trigonometric limits are a special category within limits that utilize properties of trigonometric functions. These limits often feature prominently in calculus problems involving angles and circular motion.
The key property we applied in our solution is \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \). This property is essential because it helps overcome certain indeterminate forms specific to sine and cosine functions. By pairing it with the fraction \( \frac{\sin 5x}{5x} \), we can cleverly resolve the primary expression from the exercise.
Understanding and memorizing key trigonometric limits is vital, as they provide the foundation for solving a broad array of calculus problems. In real-world applications, these trigonometric limits help model phenomena where change is inherent to angles or periodic behavior. Effectively mastering these concepts opens doors to deeper insights in both theoretical and applied mathematics.
The key property we applied in our solution is \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \). This property is essential because it helps overcome certain indeterminate forms specific to sine and cosine functions. By pairing it with the fraction \( \frac{\sin 5x}{5x} \), we can cleverly resolve the primary expression from the exercise.
Understanding and memorizing key trigonometric limits is vital, as they provide the foundation for solving a broad array of calculus problems. In real-world applications, these trigonometric limits help model phenomena where change is inherent to angles or periodic behavior. Effectively mastering these concepts opens doors to deeper insights in both theoretical and applied mathematics.
Other exercises in this chapter
Problem 39
Challenge your understanding of limits but can be evaluated using the knowledge gained in this section. $$ \lim _{x \rightarrow 0} \frac{\sin 3 x}{x} $$
View solution Problem 40
Use the Bisection Method to approximate, accurate to two decimal places, the value of the root of the given function in the given interval. $$ f(x)=\sin x-1 / 2
View solution Problem 41
Use the Bisection Method to approximate, accurate to two decimal places, the value of the root of the given function in the given interval. $$ f(x)=e^{x}-2 \tex
View solution Problem 41
Challenge your understanding of limits but can be evaluated using the knowledge gained in this section. $$ \lim _{x \rightarrow 0} \frac{\ln (1+x)}{x} $$
View solution