Problem 7
Question
Evaluate the trigonometric limits. $$ \lim _{x \rightarrow 0} \frac{\sin (5 x)}{x} $$
Step-by-Step Solution
Verified Answer
The limit is 5.
1Step 1: Identify the Standard Limit Form
The problem involves a limit that resembles a standard trigonometric limit: \( \lim _{x \to 0} \frac{\sin x}{x} = 1 \). Our goal is to manipulate the given expression to match this form or use modifications of this standard limit.
2Step 2: Rewrite the Expression
Notice that the argument of the sine function is \(5x\) instead of \(x\). To use the standard limit, rewrite \(\frac{\sin (5x)}{x}\) by multiplying and dividing by 5: \[ \frac{\sin (5x)}{x} = \frac{\sin (5x)}{5x} \cdot 5 \]
3Step 3: Apply the Standard Limit
Now, the expression \( \frac{\sin(5x)}{5x} \) matches the form of the standard limit. According to \( \lim _{x \to 0} \frac{\sin x}{x} = 1 \), we know: \( \lim _{x \to 0} \frac{\sin (5x)}{5x} = 1 \).
4Step 4: Calculate the Limit
The entire expression becomes: \[ 5 \cdot \lim _{x \to 0} \frac{\sin (5x)}{5x} = 5 \cdot 1 = 5 \] Thus, the limit is 5.
Key Concepts
Standard Limit FormLimit EvaluationSine Function Manipulation
Standard Limit Form
Understanding the concept of a standard limit form is fundamental when dealing with trigonometric limits. This involves recognizing common limit expressions where specific elements have known limits. One such critical standard limit is \( \lim _{x \to 0} \frac{\sin x}{x} = 1 \). This formula provides a simple yet powerful tool to evaluate limits involving the sine function.
Why is this important? Because it gives us a benchmark or starting point. It means that if you have a limit that can be manipulated into this form, you know its limit value without performing complex calculations. This short-cut saves time and streamlines the problem-solving process tremendously.
Standard limits help solve limits at points where substitution alone does not work. Rewriting the problem into a form that resembles these standard limits is a crucial strategy in calculus.
Why is this important? Because it gives us a benchmark or starting point. It means that if you have a limit that can be manipulated into this form, you know its limit value without performing complex calculations. This short-cut saves time and streamlines the problem-solving process tremendously.
Standard limits help solve limits at points where substitution alone does not work. Rewriting the problem into a form that resembles these standard limits is a crucial strategy in calculus.
Limit Evaluation
Limit evaluation is all about determining the value that a function approaches as the input approaches a certain point. Here, the idea is to manipulate given expressions so that they conform to known limits.
This problem revolves around transforming the initial expression \( \lim _{x \to 0} \frac{\sin(5x)}{x} \). The aim is to align it with the standard form. You achieve this by multiplying and dividing by the same term (5 in this case), leading to a workable expression:
Recognizing alignments with known standard forms is a key skill in evaluating limits efficiently. Through this method, more complex limits become simpler, making it practical to find limits using minimal steps.
This problem revolves around transforming the initial expression \( \lim _{x \to 0} \frac{\sin(5x)}{x} \). The aim is to align it with the standard form. You achieve this by multiplying and dividing by the same term (5 in this case), leading to a workable expression:
- Rewrite \( \frac{\sin(5x)}{x} \) as \( \frac{\sin(5x)}{5x} \times 5 \), which restructures the problem.
Recognizing alignments with known standard forms is a key skill in evaluating limits efficiently. Through this method, more complex limits become simpler, making it practical to find limits using minimal steps.
Sine Function Manipulation
Sine function manipulation focuses on restructuring expressions that involve the sine function in order to simplify the evaluation of limits. When we have a modest transformation, the task becomes much easier.
In the expression \( \frac{\sin(5x)}{x} \), there is a mismatch between the sine argument and the denominator. Direct comparison with \( \frac{\sin x}{x} \) requires that these expressions match. Through careful restructuring, such as multiplying and dividing by 5,
Crafting expressions into a known sine manipulation form confirms that mathematical transformations (like multiplying/dividing) can solve seemingly tricky problems. This underscores the importance of polynomial-like manipulation even for trigonometric identities.
In the expression \( \frac{\sin(5x)}{x} \), there is a mismatch between the sine argument and the denominator. Direct comparison with \( \frac{\sin x}{x} \) requires that these expressions match. Through careful restructuring, such as multiplying and dividing by 5,
- This balancing introduces \( \frac{\sin(5x)}{5x} \), compatible with the standard limit.
Crafting expressions into a known sine manipulation form confirms that mathematical transformations (like multiplying/dividing) can solve seemingly tricky problems. This underscores the importance of polynomial-like manipulation even for trigonometric identities.
Other exercises in this chapter
Problem 6
In Problems 1-32, use a table or a graph to investigate each limit. $$ \lim _{t \rightarrow \pi / 9} \sin (3 t) $$
View solution Problem 6
Evaluate the limits in problems. $$ \lim _{x \rightarrow \infty} \frac{1-5 x^{3}}{1+3 x^{4}} $$
View solution Problem 7
Let $$ f(x)=\frac{1}{x}, \quad x>0 $$ (a) Graph \(y=f(x)\) for \(0
View solution Problem 7
In Problems 1-32, use a table or a graph to investigate each limit. $$ \lim _{x \rightarrow \pi / 2} 2 \sec \frac{x}{3} $$
View solution