The given mathematical expression is:

$\lim_{x \to 0} \frac{5x}{\sin{\frac{x}{5}}}$

Solution:

We use the standard limit property:

$\lim_{u \to 0} \frac{\sin u}{u} = 1$

Let $u = \frac{x}{5}$, so as $x \to 0$, we also have $u \to 0$. Rewriting the given limit:

$\lim_{x \to 0} \frac{5x}{\sin{\frac{x}{5}}} = \lim_{x \to 0} \frac{5x}{\sin u}$

Since $u = \frac{x}{5}$, we can substitute:

$\lim_{x \to 0} \frac{5x}{\sin u} = \lim_{u \to 0} \frac{5(5u)}{\sin u}$

$= \lim_{u \to 0} 5 \cdot \frac{5u}{\sin u}$

Using the limit property $\frac{\sin u}{u} \approx 1$ as $u \to 0$, we get:

$= 5 \times 5 = 25$

Final Answer:

$\lim_{x \to 0} \frac{5x}{\sin{\frac{x}{5}}} = 25$

Would you like further clarifications or step-by-step breakdowns?

Related Questions:

1. What is the general limit property of $\frac{\sin x}{x}$ as $x \to 0$?
2. How does the substitution method work for limits?
3. What are the conditions for using L'Hôpital’s Rule in limits?
4. Can we use Taylor series to evaluate this limit?
5. What are some common trigonometric limits used in calculus?

Tip:

When dealing with limits involving trigonometric functions, always check if standard limit properties like $\lim_{x \to 0} \frac{\sin x}{x} = 1$ can simplify the expression before applying more advanced techniques.