kokoszzm

2022-06-20

Calculate the Limit as x approaches 0: $\underset{x\to 0}{lim}\frac{\mathrm{ln}\left(1+\mathrm{sin}x\right)}{\mathrm{sin}\left(2x\right)}$

laure6237ma

Hint: $\mathrm{sin}\left(2x\right)=2\mathrm{sin}x\mathrm{cos}x$ and $\frac{\mathrm{log}\left(1+\mathrm{sin}x\right)}{\mathrm{sin}x}\to 1$

Semaj Christian

Notice,
$\underset{x\to 0}{lim}\frac{\mathrm{ln}\left(1+\mathrm{sin}x\right)}{\mathrm{sin}\left(2x\right)}$
$=\underset{x\to 0}{lim}\frac{\mathrm{ln}\left(1+\mathrm{sin}x\right)}{2\mathrm{sin}x\mathrm{cos}x}$
$=\frac{1}{2}\underset{x\to 0}{lim}\frac{\mathrm{ln}\left(1+\mathrm{sin}x\right)}{\mathrm{sin}x}\cdot \frac{1}{\mathrm{cos}x}$
let $\mathrm{sin}x=t\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}t\to 0$ as $x\to 0$,
$=\frac{1}{2}\left(\underset{t\to 0}{lim}\frac{\mathrm{ln}\left(1+t\right)}{t}\right)\cdot \left(\underset{x\to 0}{lim}\frac{1}{\mathrm{cos}x}\right)$
$=\frac{1}{2}\left(1\right)\cdot \left(1\right)=\frac{1}{2}$

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