I’m confused on The limit part expression, why is there 2 sin. Why did it happen? Give a brief detail.

Certainly! Let's go through the limit step-by-step to clarify why the expression involves \(\sin^2 x\):

1. **Starting Expression:**
   The original limit is \(\lim_{x \to 0} (\sin x)(\tan x)\).

2. **Rewrite \(\tan x\):**
   We know that \(\tan x = \frac{\sin x}{\cos x}\). So, substituting this in, we get:
   \[
   \lim_{x \to 0} (\sin x)(\tan x) = \lim_{x \to 0} \left( \sin x \cdot \frac{\sin x}{\cos x} \right)
   \]

3. **Combine the Expressions:**
   When we multiply \(\sin x\) by \(\frac{\sin x}{\cos x}\), we get:
   \[
   \sin x \cdot \frac{\sin x}{\cos x} = \frac{\sin x \cdot \sin x}{\cos x} = \frac{\sin^2 x}{\cos x}
   \]
   This is why the \(\sin x\) appears squared (\(\sin^2 x\)) in the numerator of the fraction.

4. **Simplify the Limit:**
   Now the limit expression is:
   \[
   \lim_{x \to 0} \frac{\sin^2 x}{\cos x}
   \]
   To evaluate this limit, we need to consider the behavior of \(\sin x\) and \(\cos x\) as \(x\) approaches 0:
   - \(\sin x\) approaches 0.
   - \(\cos x\) approaches 1.

5. **Evaluate the Limit:**
   Therefore, substituting these limits into the expression, we get:
   \[
   \lim_{x \to 0} \frac{\sin^2 x}{\cos x} = \frac{(\lim_{x \to 0} \sin x)^2}{\lim_{x \to 0} \cos x} = \frac{0^2}{1} = 0
   \]

The key step is recognizing that \(\tan x = \frac{\sin x}{\cos x}\), which when multiplied by \(\sin x\), results in \(\frac{\sin^2 x}{\cos x}\).

Would you like any further details or have any questions?

Here are five more questions you might want to explore:
1. How do you prove the limit \(\lim_{x \to 0} \frac{\sin x}{x} = 1\)?
2. What is the limit \(\lim_{x \to 0} \frac{\cos x - 1}{x}\)?
3. How can you use series expansions to evaluate trigonometric limits?
4. What are common techniques to simplify trigonometric expressions before taking limits?
5. How do you find the limit of a function using the Squeeze Theorem?

**Tip:** When dealing with limits involving trigonometric functions, rewriting the functions in terms of sine and cosine can often simplify the evaluation process.

Explore the limit \(\lim_{x \to 0} (\sin x)(\tan x)\) and why it involves \(\sin^2 x\). Learn step-by-step how to evaluate this limit using trigonometric identities and properties. Suitable for college-level mathematics.

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