Simple Limit

**xwrathbringerx** · August 17th 2009, 04:54 PM

Lim (xà 0) [(sinx – x)/sinx]

**Random Variable** · August 17th 2009, 05:19 PM

Using L'Hospital's rule,

$\lim_{x \to 0} \frac{\sin x -x}{\sin x} = \lim_{x \to 0} \frac{\cos x -1}{\cos x} = \frac{0}{1} = 0$

or by expanding $\sin x$ as a Taylor series centered about x=0.

$\lim_{x \to 0} \frac{\sin x-x}{\sin x} = \lim_{x \to 0} \Big(1 - \frac{x}{\sin x}\Big) = \lim_{x \to 0} \Big(1 - \frac{x}{x-\frac{1}{3!}x^{3}+\frac{1}{5!}x^{5}+...}\Big)$ $= \lim_{x \to 0} \Big(1 - \frac{1}{1-\frac{1}{3!}x^{2}+\frac{1}{5!}x^{4}+...}\Big) = 1-1 =0$

**mr fantastic** · August 17th 2009, 07:45 PM

Originally Posted by xwrathbringerx

Lim (xà 0) [(sinx – x)/sinx]

Alternatively to the previous excellent solutions:

$= \lim_{x \rightarrow 0} \left( 1 - \frac{x}{\sin x} \right) = 1 - \lim_{x \rightarrow 0}\frac{x}{\sin x} = 1 -1 = 0$

using a well known standard trig limit.

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