Question

evaluate the limit using lhospitals rule
\\\lim_{x \to 0} \frac{e^x - 1}{\sin(9x)}\\

Explanation:

Verify the indeterminate form

$$ \lim_{x \to 0} (e^x - 1) = e^0 - 1 = 0 $$

$$ \lim_{x \to 0} \sin(9x) = \sin(0) = 0 $$

Apply L'Hopital's Rule

$$ \lim_{x \to 0} \frac{\frac{d}{dx}(e^x - 1)}{\frac{d}{dx}(\sin(9x))} = \lim_{x \to 0} \frac{e^x}{9\cos(9x)} $$

Evaluate the limit

$$ \frac{e^0}{9\cos(0)} = \frac{1}{9(1)} = \frac{1}{9} $$

Answer:

Evaluate the limit using L'Hospital's rule
\(\lim_{x\to0} \frac{e^x - 1}{\sin(9x)}\) = <blank>\(\frac{1}{9}\)</blank>