Limit \lim_{x\to 0} {\frac{2x-\sin{x}}{3x+\sin{x}}}

Question

Limit

lim_{x \to 0} {\frac{2 x - \sin {x}}{3 x + \sin {x}}}

Answer 1

Divide by x!

lim_{x \to 0} \frac{2 x - \sin x}{3 x + \sin x} = lim_{x \to 0} \frac{2 - \frac{\sin x}{x}}{3 + \frac{\sin x}{x}} = \frac{2 - 1}{3 + 1} = \frac{14}{}

where the well-known limit

lim_{x \to 0} \frac{\sin x}{x} = 1

was used.

Answer 2

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

Limit \lim_{x\to 0} {\frac{2x-\sin{x}}{3x+\sin{x}}}

Answers (2)

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