Question

evaluate the limit using lhospitals rule
\\\lim_{x \to 0} \frac{e^x - 1}{\sin(2x)}\\

Explanation:

Verify the indeterminate form

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

$$ \lim_{x \to 0} \sin(2x) = \sin(0) = 0 $$

Apply L'Hospital's rule

$$ \lim_{x \to 0} \frac{e^x - 1}{\sin(2x)} = \lim_{x \to 0} \frac{\frac{d}{dx}(e^x - 1)}{\frac{d}{dx}(\sin(2x))} = \lim_{x \to 0} \frac{e^x}{2\cos(2x)} $$

Evaluate the limit of the derivatives

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

Answer:

Evaluate the limit using L'Hospital's rule
\(\lim_{x \to 0} \frac{e^x - 1}{\sin (2x)}\) = <blank>\(\frac{1}{2}\)</blank>