lim_(x rarr0)(sin 3x)/(sin 2x)=

Identify Limit: Identify the limit to be evaluated. We need to find the limit of the function (sin 3x)/(sin 2x) as x approaches 0.Apply Limit: Apply the limit to the function. lim_(x rarr0)(sin 3x)/(sin 2x) We will use the fact that lim_(x rarr0)(sin x)/x = 1, which is a well-known trigonometric limit.Rewrite Function: Rewrite the function to use the known limit. We can rewrite the function as (3/2) * (sin 3x)/(3x) * (2x)/(sin 2x).Apply Known Limit: Apply the limit to each part of the function. lim_(x rarr0)(3/2) * (sin 3x)/(3x) * (2x)/(sin 2x) = (3/2) * lim_(x rarr0)(sin 3x)/(3x) * lim_(x rarr0)(2x)/(sin 2x)Evaluate Limits: Evaluate each limit separately. lim_(x rarr0)(sin 3x)/(3x) = 1 and lim_(x rarr0)(2x)/(sin 2x) = 1, using the known trigonometric limit.Multiply Results: Multiply the results of the limits. (3/2) * 1 * 1 = 3/2Conclude Answer: Conclude the final answer. The limit of (sin 3x)/(sin 2x) as x approaches 0 is 3/2.

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Math Problems

Calculus

Find limits involving trigonometric functions

Full solution

\lim_{x \to 0}\left(\frac{\sin 3x}{\sin 2x}\right)= \square

Identify Limit: Identify the limit to be evaluated. $\newline$ We need to find the limit of the function $\frac{\sin 3x}{\sin 2x}$ as $x$ approaches $0$ .
Apply Limit: Apply the limit to the function. $\newline$ $\lim_{x \rightarrow 0}\frac{\sin 3x}{\sin 2x}$ $\newline$ We will use the fact that $\lim_{x \rightarrow 0}\frac{\sin x}{x} = 1$ , which is a well-known trigonometric limit.
Rewrite Function: Rewrite the function to use the known limit. $\newline$ We can rewrite the function as $(\frac{3}{2}) \cdot (\frac{\sin 3x}{3x}) \cdot (\frac{2x}{\sin 2x})$ .
Apply Known Limit: Apply the limit to each part of the function. $\newline$ $\lim_{x \to 0}\left(\frac{3}{2}\right) \cdot \frac{\sin 3x}{3x} \cdot \frac{2x}{\sin 2x} = \left(\frac{3}{2}\right) \cdot \lim_{x \to 0}\frac{\sin 3x}{3x} \cdot \lim_{x \to 0}\frac{2x}{\sin 2x}$
Evaluate Limits: Evaluate each limit separately. $\newline$ $\lim_{x \rightarrow 0}\frac{\sin 3x}{3x} = 1$ and $\lim_{x \rightarrow 0}\frac{2x}{\sin 2x} = 1$ , using the known trigonometric limit.
Multiply Results: Multiply the results of the limits. $\newline$ $(\frac{3}{2}) \times 1 \times 1 = \frac{3}{2}$
Conclude Answer: Conclude the final answer. $\newline$ The limit of $\frac{\sin 3x}{\sin 2x}$ as $x$ approaches $0$ is $\frac{3}{2}$ .