To evaluate the limit using L'Hôpital's Rule, let's proceed with the given expression:

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

Step 1: Verify the Indeterminate Form

As $x \to 0$:

• The numerator $\sin 5x \to 0$,
• The denominator $x \to 0$.

This creates the indeterminate form $\frac{0}{0}$, which allows us to apply L'Hôpital's Rule.

Step 2: Apply L'Hôpital's Rule

According to L'Hôpital's Rule, if we have an indeterminate form $\frac{0}{0}$, we can differentiate the numerator and the denominator separately:

1. Differentiate the numerator: $\frac{d}{dx}(\sin 5x) = 5 \cos 5x$.
2. Differentiate the denominator: $\frac{d}{dx}(x) = 1$.

The limit then becomes:

$\lim_{{x \to 0}} \frac{5 \cos 5x}{1}$

Step 3: Substitute $x = 0$

Now, substitute $x = 0$:

$5 \cos 5(0) = 5 \cos 0 = 5 \times 1 = 5$

Final Answer

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

Would you like further clarification or details on any of the steps?

Related Questions:

1. How would the limit change if we evaluated $\lim_{{x \to 0}} \frac{\sin kx}{x}$ for any constant $k$?
2. Can we apply L'Hôpital's Rule to solve $\lim_{{x \to 0}} \frac{\sin x}{x^2}$?
3. What is the result of $\lim_{{x \to 0}} \frac{\tan 5x}{x}$?
4. Why is L'Hôpital's Rule applicable only in certain indeterminate forms?
5. How does the result differ if we calculate $\lim_{{x \to 0}} \frac{\sin 5x}{2x}$?

Tip:

When using L'Hôpital's Rule, always verify that the limit initially presents an indeterminate form like $\frac{0}{0}$ or $\frac{\infty}{\infty}$ before proceeding.