Question

evaluate the limit using lhospitals rule if necessary
\\\lim_{x \to 0} \frac{\sin(12x)}{\sin(15x)}\\

Explanation:

Verify the indeterminate form

$$ \lim_{x \to 0} \sin(12x) = \sin(0) = 0 $$

$$ \lim_{x \to 0} \sin(15x) = \sin(0) = 0 $$

Since the limit yields the indeterminate form \(\frac{0}{0}\), L'Hôpital's rule is applicable.

Apply L'Hôpital's rule

$$ \lim_{x \to 0} \frac{\sin(12x)}{\sin(15x)} = \lim_{x \to 0} \frac{\frac{d}{dx}[\sin(12x)]}{\frac{d}{dx}[\sin(15x)]} = \lim_{x \to 0} \frac{12\cos(12x)}{15\cos(15x)} $$

Evaluate the limit of the derivatives

$$ \lim_{x \to 0} \frac{12\cos(12x)}{15\cos(15x)} = \frac{12\cos(0)}{15\cos(0)} = \frac{12(1)}{15(1)} = \frac{12}{15} = \frac{4}{5} $$

Answer:

Evaluate the limit using L'Hospital's rule if necessary
\(\lim_{x\to0} \frac{\sin(12x)}{\sin(15x)}\) = <blank>\(\frac{4}{5}\)</blank>