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To find the limit of [tex]\(\frac{\tan x}{x}\)[/tex] as [tex]\(x\)[/tex] approaches 0, we will go through a step-by-step analysis.
1. Understand the Function: The function we are dealing with is [tex]\(\frac{\tan x}{x}\)[/tex]. As [tex]\(x\)[/tex] approaches 0, both the numerator ([tex]\(\tan x\)[/tex]) and the denominator ([tex]\(x\)[/tex]) approach 0, which gives us an indeterminate form [tex]\(\frac{0}{0}\)[/tex]. This means we need to manipulate the function or use limit properties to find the value.
2. L'Hôpital's Rule: One method to solve limits of the form [tex]\(\frac{0}{0}\)[/tex] is L'Hôpital's Rule, which states that:
[tex]\[ \lim_{x \to 0} \frac{f(x)}{g(x)} = \lim_{x \to 0} \frac{f'(x)}{g'(x)} \][/tex]
if the limit on the right-hand side exists. Here, [tex]\(f(x) = \tan x\)[/tex] and [tex]\(g(x) = x\)[/tex].
3. Differentiate the Numerator and the Denominator:
- The derivative of [tex]\(\tan x\)[/tex] is [tex]\(\sec^2 x\)[/tex].
- The derivative of [tex]\(x\)[/tex] is [tex]\(1\)[/tex].
4. Apply L'Hôpital's Rule:
[tex]\[ \lim_{x \to 0} \frac{\tan x}{x} = \lim_{x \to 0} \frac{\sec^2 x}{1} \][/tex]
5. Evaluate the Limit:
- We know that [tex]\(\sec^2 x = 1 + \tan^2 x\)[/tex], and as [tex]\(x\)[/tex] approaches 0, [tex]\(\tan x\)[/tex] also approaches 0.
- Thus, [tex]\(\sec^2(0) = 1\)[/tex].
Therefore,
[tex]\[ \lim_{x \to 0} \frac{\sec^2 x}{1} = \sec^2(0) = 1 \][/tex]
Hence, the limit is:
[tex]\[ \lim _{x \rightarrow 0} \frac{\tan x}{x} = 1 \][/tex]
1. Understand the Function: The function we are dealing with is [tex]\(\frac{\tan x}{x}\)[/tex]. As [tex]\(x\)[/tex] approaches 0, both the numerator ([tex]\(\tan x\)[/tex]) and the denominator ([tex]\(x\)[/tex]) approach 0, which gives us an indeterminate form [tex]\(\frac{0}{0}\)[/tex]. This means we need to manipulate the function or use limit properties to find the value.
2. L'Hôpital's Rule: One method to solve limits of the form [tex]\(\frac{0}{0}\)[/tex] is L'Hôpital's Rule, which states that:
[tex]\[ \lim_{x \to 0} \frac{f(x)}{g(x)} = \lim_{x \to 0} \frac{f'(x)}{g'(x)} \][/tex]
if the limit on the right-hand side exists. Here, [tex]\(f(x) = \tan x\)[/tex] and [tex]\(g(x) = x\)[/tex].
3. Differentiate the Numerator and the Denominator:
- The derivative of [tex]\(\tan x\)[/tex] is [tex]\(\sec^2 x\)[/tex].
- The derivative of [tex]\(x\)[/tex] is [tex]\(1\)[/tex].
4. Apply L'Hôpital's Rule:
[tex]\[ \lim_{x \to 0} \frac{\tan x}{x} = \lim_{x \to 0} \frac{\sec^2 x}{1} \][/tex]
5. Evaluate the Limit:
- We know that [tex]\(\sec^2 x = 1 + \tan^2 x\)[/tex], and as [tex]\(x\)[/tex] approaches 0, [tex]\(\tan x\)[/tex] also approaches 0.
- Thus, [tex]\(\sec^2(0) = 1\)[/tex].
Therefore,
[tex]\[ \lim_{x \to 0} \frac{\sec^2 x}{1} = \sec^2(0) = 1 \][/tex]
Hence, the limit is:
[tex]\[ \lim _{x \rightarrow 0} \frac{\tan x}{x} = 1 \][/tex]
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