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.