How do you find the limit of $\frac{x + sin 2 x}{3 x}$ $( x+sin 2x)/( 3x)$ as x approaches 0?

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How do you find the limit of $\frac{x + sin 2 x}{3 x}$ $( x+sin 2x)/( 3x)$ as x approaches 0?

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Answer 1 · 2016-07-01T15:58:02

$L t_{x \to 0} \frac{x + sin 2 x}{3 x} = 1$ $Lt_(x->0)(x+sin2x)/(3x)=1$

Explanation:

$L t_{x \to 0} \frac{x + sin 2 x}{3 x}$ $Lt_(x->0)(x+sin2x)/(3x)$

= $L t_{x \to 0} (\frac{x}{3 x} + \frac{sin 2 x}{3 x})$ $Lt_(x->0)(x/(3x)+(sin2x)/(3x))$

= $L t_{x \to 0} (\frac{1}{3} + \frac{sin 2 x}{2 x} \times \frac{2 x}{3 x})$ $Lt_(x->0)(1/3+(sin2x)/(2x)xx(2x)/(3x))$

= $L t_{x \to 0} (\frac{1}{3}) + L t_{x \to 0} \frac{sin 2 x}{2 x} \times L t_{x \to 0} \frac{2 x}{3 x}$ $Lt_(x->0)(1/3)+Lt_(x->0)(sin2x)/(2x)xxLt_(x->0)(2x)/(3x)$

= $\frac{1}{3} + L t_{2 x \to 0} \frac{sin 2 x}{2 x} \times \frac{2}{3}$ $1/3+Lt_(2x->0)(sin2x)/(2x)xx2/3$

= $\frac{1}{3} + 1 \times \frac{2}{3}$ $1/3+1xx2/3$

= $\frac{1}{3} + \frac{2}{3} = 1$ $1/3+2/3=1$

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How do you find the limit of $\frac{x + sin 2 x}{3 x}$ $( x+sin 2x)/( 3x)$ as x approaches 0?

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Explanation:

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How do you find the limit of $\frac{x + sin 2 x}{3 x}$ $( x+sin 2x)/( 3x)$ as x approaches 0?