Connect with a community of experts and enthusiasts on IDNLearn.com. Discover in-depth and reliable answers to all your questions from our knowledgeable community members who are always ready to assist.
Sagot :
To determine the limit of [tex]\(\frac{\sin x}{x}\)[/tex] as [tex]\(x\)[/tex] approaches 0, we need to evaluate the expression within the context of limits in calculus. Here is a detailed, step-by-step solution to this problem:
1. Understanding the Expression:
We are given the limit expression [tex]\(\lim_{x \to 0} \frac{\sin x}{x}\)[/tex]. This is a classic limit in calculus and is often taught because it has fundamental importance in understanding the behavior of trigonometric functions around zero.
2. Behavior of the Function Near Zero:
As [tex]\(x\)[/tex] approaches 0, [tex]\(\sin x\)[/tex] also approaches 0 because [tex]\(\sin x\)[/tex] is continuous and smooth at zero. Naively substituting [tex]\(x = 0\)[/tex] directly would result in the form [tex]\(\frac{0}{0}\)[/tex], which is indeterminate. This indicates that we need to evaluate the limit using other techniques.
3. Taylor Series Expansion:
One way to understand this limit is by using the Taylor series expansion of [tex]\(\sin x\)[/tex] around [tex]\(x = 0\)[/tex].
The Taylor series for [tex]\(\sin x\)[/tex] is:
[tex]\[ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \][/tex]
For small values of [tex]\(x\)[/tex], we can approximate [tex]\(\sin x \approx x\)[/tex]. Substituting this approximation in the given limit, we have:
[tex]\[ \frac{\sin x}{x} \approx \frac{x}{x} = 1 \][/tex]
Hence, as [tex]\(x\)[/tex] approaches 0, [tex]\(\frac{\sin x}{x}\)[/tex] approaches 1.
4. Squeeze Theorem:
Another way is to use the Squeeze Theorem. For [tex]\(x \neq 0\)[/tex], we know the following inequalities hold:
[tex]\[ \cos x \leq \frac{\sin x}{x} \leq 1 \][/tex]
Near [tex]\(x = 0\)[/tex], [tex]\(\cos x\)[/tex] is very close to 1 because [tex]\(\cos(0) = 1\)[/tex]. By the Squeeze Theorem:
[tex]\[ \lim_{x \to 0} \cos x = 1 \quad \text{and} \quad \lim_{x \to 0} 1 = 1 \][/tex]
Therefore,
[tex]\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \][/tex]
5. Graphical Insight:
Graphing the function [tex]\(\frac{\sin x}{x}\)[/tex] around [tex]\(x = 0\)[/tex] supports our findings. The graph of [tex]\(\frac{\sin x}{x}\)[/tex] shows that as [tex]\(x\)[/tex] gets closer to 0 from either the positive or negative side, the function's value approaches 1.
Considering these various approaches, we can confirm that the limit of [tex]\(\frac{\sin x}{x}\)[/tex] as [tex]\(x\)[/tex] approaches 0 is:
[tex]\[ \boxed{1} \][/tex]
1. Understanding the Expression:
We are given the limit expression [tex]\(\lim_{x \to 0} \frac{\sin x}{x}\)[/tex]. This is a classic limit in calculus and is often taught because it has fundamental importance in understanding the behavior of trigonometric functions around zero.
2. Behavior of the Function Near Zero:
As [tex]\(x\)[/tex] approaches 0, [tex]\(\sin x\)[/tex] also approaches 0 because [tex]\(\sin x\)[/tex] is continuous and smooth at zero. Naively substituting [tex]\(x = 0\)[/tex] directly would result in the form [tex]\(\frac{0}{0}\)[/tex], which is indeterminate. This indicates that we need to evaluate the limit using other techniques.
3. Taylor Series Expansion:
One way to understand this limit is by using the Taylor series expansion of [tex]\(\sin x\)[/tex] around [tex]\(x = 0\)[/tex].
The Taylor series for [tex]\(\sin x\)[/tex] is:
[tex]\[ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \][/tex]
For small values of [tex]\(x\)[/tex], we can approximate [tex]\(\sin x \approx x\)[/tex]. Substituting this approximation in the given limit, we have:
[tex]\[ \frac{\sin x}{x} \approx \frac{x}{x} = 1 \][/tex]
Hence, as [tex]\(x\)[/tex] approaches 0, [tex]\(\frac{\sin x}{x}\)[/tex] approaches 1.
4. Squeeze Theorem:
Another way is to use the Squeeze Theorem. For [tex]\(x \neq 0\)[/tex], we know the following inequalities hold:
[tex]\[ \cos x \leq \frac{\sin x}{x} \leq 1 \][/tex]
Near [tex]\(x = 0\)[/tex], [tex]\(\cos x\)[/tex] is very close to 1 because [tex]\(\cos(0) = 1\)[/tex]. By the Squeeze Theorem:
[tex]\[ \lim_{x \to 0} \cos x = 1 \quad \text{and} \quad \lim_{x \to 0} 1 = 1 \][/tex]
Therefore,
[tex]\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \][/tex]
5. Graphical Insight:
Graphing the function [tex]\(\frac{\sin x}{x}\)[/tex] around [tex]\(x = 0\)[/tex] supports our findings. The graph of [tex]\(\frac{\sin x}{x}\)[/tex] shows that as [tex]\(x\)[/tex] gets closer to 0 from either the positive or negative side, the function's value approaches 1.
Considering these various approaches, we can confirm that the limit of [tex]\(\frac{\sin x}{x}\)[/tex] as [tex]\(x\)[/tex] approaches 0 is:
[tex]\[ \boxed{1} \][/tex]
We appreciate your contributions to this forum. Don't forget to check back for the latest answers. Keep asking, answering, and sharing useful information. For dependable answers, trust IDNLearn.com. Thank you for visiting, and we look forward to assisting you again.