\lim_{x\to{0}}{\frac{1-\sin\left(x\right)}{\tan\left(x\right)}}

We are asked to find the limit:

\[
\lim_{x \to 0} \frac{1 - \sin(x)}{\tan(x)}
\]

### Step 1: Rewrite \(\tan(x)\)
We know that \(\tan(x) = \frac{\sin(x)}{\cos(x)}\). Substituting this into the original expression:

\[
\lim_{x \to 0} \frac{1 - \sin(x)}{\frac{\sin(x)}{\cos(x)}} = \lim_{x \to 0} \frac{(1 - \sin(x)) \cdot \cos(x)}{\sin(x)}
\]

### Step 2: Apply L'Hôpital's Rule (optional)
At \(x = 0\), both the numerator and denominator tend to 0, which allows us to apply **L'Hôpital's Rule**. We need to differentiate both the numerator and denominator with respect to \(x\).

- Derivative of the numerator \(1 - \sin(x)\):
\[
\frac{d}{dx}(1 - \sin(x)) = -\cos(x)
\]

- Derivative of the denominator \(\sin(x)\):
\[
\frac{d}{dx}(\sin(x)) = \cos(x)
\]

Applying L'Hôpital's Rule:

\[
\lim_{x \to 0} \frac{-\cos(x) \cdot \cos(x)}{\cos(x)} = \lim_{x \to 0} -\cos(x)
\]

### Step 3: Evaluate the limit
As \(x \to 0\), \(\cos(0) = 1\). Thus, we get:

\[
\lim_{x \to 0} -\cos(x) = -1
\]

### Final Answer:
\[
\boxed{-1}
\]

Would you like further details on any of the steps? Here are some related questions:

1. How do we apply L'Hôpital's rule to indeterminate forms?
2. What are the conditions for using L'Hôpital's Rule?
3. Can we solve the limit without L'Hôpital's Rule using Taylor series?
4. Why does \(\tan(x) \approx x\) near \(x = 0\)?
5. How does the sine function behave as \(x \to 0\)?

**Tip:** When both the numerator and denominator approach 0, it's often a sign that L'Hôpital's Rule may be useful.

Explore how to solve the limit problem lim (1 - sin(x)) / tan(x) as x approaches 0 using L'Hôpital's Rule. This detailed solution covers key calculus concepts such as trigonometric identities and differentiation. Suitable for Grades 11-12 and college-level calculus students.

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