Anika Boyd
2022-03-22
Answered

Limit $\underset{x\to 0}{lim}\left\{\frac{2x-\mathrm{sin}\left\{x\right\}}{3x+\mathrm{sin}\left\{x\right\}}\right\}$

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Aarlenlsi1

Answered 2022-03-23
Author has **10** answers

Divide by x!

$\underset{x\to 0}{lim}\frac{2x-\mathrm{sin}x}{3x+\mathrm{sin}x}=\underset{x\to 0}{lim}\frac{2-\frac{\mathrm{sin}x}{x}}{3+\frac{\mathrm{sin}x}{x}}=\frac{2-1}{3+1}=\frac{14}{}$

where the well-known limit$\underset{x\to 0}{lim}\frac{\mathrm{sin}x}{x}=1$ was used.

where the well-known limit

Mercedes Chang

Answered 2022-03-24
Author has **15** answers

Since $\mathrm{sin}x=x+O\left(x\right)$ for $x\to 0$ ,your limit becomes: $\underset{x\to 0}{lim}\frac{2x-x+O\left({x}^{3}\right)}{3x+x-O\left({x}^{3}\right)}$ thence for $x\to 0$ the result is $\frac{1}{4}$

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