Questions: Find the limit. (If an answer does not exist, enter DNE.) lim x -> 0 (sin 3x / x)

Solution Steps

To find the limit of \(\lim _{x \rightarrow 0} \frac{\sin 3x}{x}\), we can use the standard limit property \(\lim_{x \to 0} \frac{\sin x}{x} = 1\). By making a substitution, we can express the given limit in a form that allows us to apply this property.

1. Recognize that the expression \(\frac{\sin 3x}{x}\) can be rewritten as \(\frac{\sin 3x}{3x} \cdot 3\).
2. Use the limit property \(\lim_{u \to 0} \frac{\sin u}{u} = 1\) by letting \(u = 3x\).
3. Evaluate the limit as \(x\) approaches 0.

To evaluate the limit \(\lim_{x \to 0} \frac{\sin 3x}{x}\), we can rewrite the expression as \(\frac{\sin 3x}{3x} \cdot 3\). This allows us to use the standard limit property.

We know that \(\lim_{u \to 0} \frac{\sin u}{u} = 1\). By letting \(u = 3x\), we have: \[ \lim_{x \to 0} \frac{\sin 3x}{3x} = 1 \]

Using the rewritten expression \(\frac{\sin 3x}{3x} \cdot 3\), we can evaluate the limit: \[ \lim_{x \to 0} \left(\frac{\sin 3x}{3x} \cdot 3\right) = 1 \cdot 3 = 3 \]

Final Answer