To evaluate the limit

$\lim_{x \to 0} \frac{5x - 5x \cos(x)}{7x \sin(x)},$

let’s simplify the expression step-by-step.

Step 1: Factor out $x$ in the numerator and denominator

Rewrite the expression by factoring $x$ from both the numerator and the denominator:

$\frac{5x - 5x \cos(x)}{7x \sin(x)} = \frac{5x(1 - \cos(x))}{7x \sin(x)}.$

Now, cancel $x$ from the numerator and the denominator:

$= \frac{5(1 - \cos(x))}{7 \sin(x)}.$

Step 2: Apply limit properties and trigonometric identities

Now we can directly substitute the limit $x \to 0$:

1. For $1 - \cos(x)$: We know that $1 - \cos(x) \approx \frac{x^2}{2}$ as $x \to 0$.
2. For $\sin(x)$: We also know that $\sin(x) \approx x$ as $x \to 0$.

Using these approximations, substitute into the fraction:

$\frac{5 \cdot \frac{x^2}{2}}{7 \cdot x} = \frac{\frac{5x^2}{2}}{7x}.$

Step 3: Simplify the fraction

Now, divide both the numerator and denominator by $x$:

$= \frac{\frac{5x}{2}}{7} = \frac{5x}{14}.$

Step 4: Take the limit as $x \to 0$

As $x \to 0$, the term $\frac{5x}{14} \to 0$.

Conclusion

Thus,

$\lim_{x \to 0} \frac{5x - 5x \cos(x)}{7x \sin(x)} = 0.$

Would you like more details on each step or have any questions?

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Related Questions

1. How do trigonometric approximations like $\sin(x) \approx x$ help in limit problems?
2. What happens if we use L'Hôpital's Rule to solve this limit?
3. Can we use series expansions for more complex trigonometric limits?
4. Why does $1 - \cos(x) \approx \frac{x^2}{2}$ as $x \to 0$?
5. How can this problem be adapted to use polar coordinates?

Tip

In trigonometric limits, approximations are often essential. Familiarize yourself with common ones like $\sin(x) \approx x$ and $1 - \cos(x) \approx \frac{x^2}{2}$ for faster problem-solving.