Evaluate the limit: limit, start subscript, x, right arrow, 0, end subscript, start fraction, 5, x, minus, 5, x, cosine, left parenthesis, x, right parenthesis, divided by, 7, x, sine, left parenthesis, x, right parenthesis, end fraction
x→0
lim
​

7xsin(x)
5x−5xcos(x)
​

To evaluate the limit

\[
\lim_{{x \to 0}} \frac{5x - 5x \cos(x)}{7x \sin(x)}
\]

we can simplify and apply L'Hôpital's Rule if necessary. Here are the steps:

### Step 1: Simplify the Expression
Rewrite the numerator:
\[
5x - 5x \cos(x) = 5x (1 - \cos(x))
\]
so the limit becomes
\[
\lim_{{x \to 0}} \frac{5x (1 - \cos(x))}{7x \sin(x)}.
\]

### Step 2: Cancel \( x \) Terms
Since \( x \neq 0 \) for the limit process, we can cancel \( x \) in the numerator and denominator:
\[
= \lim_{{x \to 0}} \frac{5 (1 - \cos(x))}{7 \sin(x)}.
\]

### Step 3: Apply Known Limits
Using the fact that \(\lim_{{x \to 0}} \frac{1 - \cos(x)}{x^2} = \frac{1}{2}\) and \(\lim_{{x \to 0}} \frac{\sin(x)}{x} = 1\), rewrite \(1 - \cos(x)\) in terms of \(x^2\) near zero.

Rewrite the limit as:
\[
= \lim_{{x \to 0}} \frac{5 \cdot \frac{1 - \cos(x)}{x^2}}{7 \cdot \frac{\sin(x)}{x}} \cdot x.
\]

Then substitute the known limits:
\[
= \frac{5 \cdot \frac{1}{2}}{7 \cdot 1} = \frac{5}{14}.
\]

### Final Answer
\[
\lim_{{x \to 0}} \frac{5x - 5x \cos(x)}{7x \sin(x)} = \frac{5}{14}.
\]

Would you like more detailed steps on any part of this? 

Here are some related questions:
1. How can L'Hôpital's Rule be used to solve similar limits?
2. What is the Taylor expansion of \(\cos(x)\) around \(x = 0\)?
3. How does the small-angle approximation for sine and cosine functions help in limits?
4. Why does \(\frac{1 - \cos(x)}{x^2} \approx \frac{1}{2}\) as \(x \to 0\)?
5. How do we know when to cancel terms in a limit problem?

**Tip:** Memorizing key trigonometric limits can make solving these problems much faster.

Evaluate the limit: lim_{x→0} (5x - 5x*cos(x)) / (7x*sin(x))

Learn how to evaluate the limit lim_{x→0} (5x - 5x*cos(x)) / (7x*sin(x)) with a detailed step-by-step solution. This problem involves key concepts in limits and trigonometric identities, using L'Hôpital's Rule and known trigonometric limits to reach the final answer of 5/14. Suitable for advanced high school students.

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