Question

evaluate the limit
\\\lim_{x \to 0} \frac{\sin 6x}{\sin 3x}\\

Explanation:

Identify the limit form

Using the Trigonometric Limits knowledge point

$$ \lim_{x \to 0} \sin 6x = \sin(0) = 0 $$

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

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

Apply L'Hopital's Rule

Using the LHopitals Rule knowledge point

$$ \lim_{x \to 0} \frac{\sin 6x}{\sin 3x} = \lim_{x \to 0} \frac{\frac{d}{dx}(\sin 6x)}{\frac{d}{dx}(\sin 3x)} = \lim_{x \to 0} \frac{6 \cos 6x}{3 \cos 3x} $$

Evaluate the simplified limit

Using the Trigonometric Limits knowledge point

$$ \lim_{x \to 0} \frac{6 \cos 6x}{3 \cos 3x} = \frac{6 \cos(0)}{3 \cos(0)} = \frac{6(1)}{3(1)} = 2 $$

Answer:

Evaluate the limit

$$\lim_{x\to0} \frac{\sin 6x}{\sin 3x}$$

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