Evaluate the limit of tan(2x)/sin(3x) as x approaches 0.

Step-by-step solution

1. 1

   Check the form. Both $\tan 2x$ and $\sin 3x$ vanish at $x=0$, so the quotient is $\frac{0}{0}$.

2. 2

   Rewrite the tangent. Use $\tan 2x=\frac{\sin 2x}{\cos 2x}$, turning the expression into $\frac{\sin 2x}{\cos 2x\,\sin 3x}$. The cosine is harmless because $\cos 0=1\neq 0$.

3. 3

   Divide numerator and denominator by $x$. The quotient becomes $\frac{\frac{\sin 2x}{x}}{\cos 2x\cdot\frac{\sin 3x}{x}}$, so every trigonometric piece is now in the standard $\frac{\sin kx}{x}$ shape.

4. 4

   Apply the standard limit. With $\lim_{x\to 0}\frac{\sin kx}{x}=k$, the numerator tends to $2$ and the factor $\frac{\sin 3x}{x}$ tends to $3$, while $\cos 2x\to 1$.

5. 5

   Combine. The limit is $\frac{2}{1\cdot 3}=\frac{2}{3}$.

6. 6

   Numerical check. At $x=10^{-6}$ the original quotient evaluates to $0.66666667$, agreeing with $\frac{2}{3}$.