Answer :
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]
