 # limit \lim_{t \to 0}\frac{\sin 3t}{\tan2t} I am struggling with this question. Judith Warner 2022-04-29 Answered
limit $\underset{t\to 0}{lim}\frac{\mathrm{sin}3t}{\mathrm{tan}2t}$
I am struggling with this question. I have attempted it, but I keep getting stuck on the following step:
$\underset{t\to 0}{lim}\frac{1}{2}\frac{\mathrm{sin}3t\mathrm{cos}2t}{\mathrm{sin}t\mathrm{cos}t}$
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You started correctly. Now multiply both numerator and denominator by corresponding ts to get:
$L=\underset{t\to 0}{lim}\frac{\mathrm{sin}\left\{3t\right\}}{3t}\cdot \frac{2t}{\mathrm{sin}\left\{2t\right\}}\cdot \frac{3}{2}\cdot \mathrm{cos}\left\{2t\right\}=1\cdot 1\cdot \frac{3}{2}\cdot 1=\frac{3}{2}.$
Note: It was used the remarkable limit: $\underset{x\to 0}{lim}\frac{\mathrm{sin}\left\{x\right\}}{x}=\underset{x\to 0}{lim}\frac{x}{\mathrm{sin}\left\{x\right\}}=1$
###### Not exactly what you’re looking for? hadnya1qd
I suspect the denominateur is $\mathrm{tan}2t$. If I'm right, just use equivalents: near 0, we have
$\mathrm{sin}u{\text{~}}_{0}u,\phantom{\rule{1em}{0ex}}\mathrm{tan}u{\text{~}}_{0}u$
so that
$\frac{\mathrm{sin}3x}{\mathrm{tan}2x}{\text{~}}_{0}\frac{3x}{2x}=\frac{3}{2}$