Question

evaluate the limit using lhospitals rule
\\\lim_{x \to 0} \frac{\sin(8x)}{\tan(3x)}\\

Explanation:

Verify the indeterminate form

$$ \lim_{x \to 0} \sin(8x) = \sin(0) = 0 $$

$$ \lim_{x \to 0} \tan(3x) = \tan(0) = 0 $$

The limit is of the indeterminate form \(\frac{0}{0}\).

Apply L'Hôpital's Rule

$$ \lim_{x \to 0} \frac{\sin(8x)}{\tan(3x)} = \lim_{x \to 0} \frac{\frac{d}{dx}[\sin(8x)]}{\frac{d}{dx}[\tan(3x)]} $$

$$ \frac{d}{dx}[\sin(8x)] = 8\cos(8x) $$

$$ \frac{d}{dx}[\tan(3x)] = 3\sec^2(3x) $$

Evaluate the limit of the derivatives

$$ \lim_{x \to 0} \frac{8\cos(8x)}{3\sec^2(3x)} = \frac{8\cos(0)}{3\sec^2(0)} = \frac{8(1)}{3(1)^2} = \frac{8}{3} $$

Answer:

Evaluate the limit using L'Hospital's rule
\(\lim_{x \to 0} \frac{\sin(8x)}{\tan(3x)} =\) <blank>\(\frac{8}{3}\)</blank>