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$$\lim_{x\to0}\frac{\ 3x + 2 - 2 \cos x}{6\sin x}$$

By looking at its graph, the limit is $1/2$.

I tried to factor and manipulate the function to be in the form that will give me the value $1/2$ but I didn't get it.

I know that $\lim_{x\to0}\frac{\ 3x }{6\sin x} = 1/2$

But how to deal the rest ?

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$$\frac{1-\cos x}{\sin x}=\frac{2\sin^2x/2}{2\sin x/2\cos x/2}=\tan x/2$$

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  • $\begingroup$ Thanks :) .. this is what i was looking for .. $\endgroup$ – Maher Oct 17 '14 at 17:22
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Applying L'Hôpitals Rule:

$$\lim_{x\to0}\frac{3x+2-2\cos x}{6\sin x}=\lim_{x\to0}\frac{3+2\sin x}{6\cos x}=\frac36=\frac12$$

Alternatively, note that in the neighbourhood of $x=0$ $\sin x\sim x$ and $\cos x\sim 1$, so:

$$\lim_{x\to0}\frac{3x+2-2\cos x}{6\sin x}=\lim_{x\to 0}\frac{3x+2-2}{6x}=\lim_{x\to0}\frac36=\frac12$$

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  • $\begingroup$ is there a way different from L'Hôpitals Rule to solve this problem? $\endgroup$ – Maher Oct 17 '14 at 13:57
  • $\begingroup$ @Maher, edited the answer. $\endgroup$ – rae306 Oct 17 '14 at 14:05
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There are three methods to the question:

$L'Hôpitals Rule:$

$$\eqalign{ & \mathop {\lim }\limits_{x \to 0} {{3x + 2 - 2\cos x} \over {6\sin x}} \cr & = \mathop {\lim }\limits_{x \to 0} {{3 + 2\sin x} \over {6\cos x}} \cr & = {1 \over 2} \cr} $$

Applying $\mathop {\lim }\limits_{x \to {x_0}} \left( {f(x) + g(x)} \right) = \mathop {\lim }\limits_{x \to {x_0}} f(x) + \mathop {\lim }\limits_{x \to {x_0}} g(x) ~~\& ~~1-cosx \sim \frac{\displaystyle 1}{\displaystyle 2}x^2:$

( $\mathop {\lim }\limits_{x \to {x_0}} f(x)$ and $\mathop {\lim }\limits_{x \to {x_0}} g(x)$ exist )

$$\eqalign{ & \mathop {\lim }\limits_{x \to 0} {{3x + 2 - 2\cos x} \over {6\sin x}} \cr & = \mathop {\lim }\limits_{x \to 0} {{3x} \over {6\sin x}} + \mathop {\lim }\limits_{x \to 0} {{2(1 - \cos x)} \over {6\sin x}} \cr & = {1 \over 2} + \mathop {\lim }\limits_{x \to 0} {{2 \cdot {1 \over 2}{x^2}} \over {6x}} \cr & = {1 \over 2} \cr} $$

$Taylor ~ Expansion:$

$$\eqalign{ & \mathop {\lim }\limits_{x \to 0} {{3x + 2 - 2\cos x} \over {6\sin x}} \cr & = \mathop {\lim }\limits_{x \to 0} {{3x + 2 - 2\left( {1 - \frac{\displaystyle 1}{\displaystyle 2}{x^2} + o({x^2})} \right)} \over {6\sin x}} \cr & = \mathop {\lim }\limits_{x \to 0} {{3x} \over {6\sin x}} + \mathop {\lim }\limits_{x \to 0} {{{x^2}} \over {6\sin x}} - 2\mathop {\lim }\limits_{x \to 0} {{o({x^2})} \over {6\sin x}} \cr & = {1 \over 2} + 0 - 0 \cr & = {1 \over 2} \cr} $$

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  • $\begingroup$ 2) et 3) are quite similar. $\endgroup$ – anderstood Oct 17 '14 at 14:29
  • $\begingroup$ @anderstood yeah. But in most cases we'd better choose 3) for that either $\mathop {\lim }\limits_{x \to {x_0}} f(x)$ or $\mathop {\lim }\limits_{x \to {x_0}} g(x)$ DNE in some cases $\endgroup$ – Shine Mic Oct 17 '14 at 14:34
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    $\begingroup$ In both 2) and 3) you suppose that the limits exist (because in both cases you split the limit of the sum), and in 2) the "~" can be proven from the Taylor expansion of $\cos(x)$. So solution 2) does not bring much, I think. Michael's solution is a "real" other method :) $\endgroup$ – anderstood Oct 17 '14 at 14:47
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$\frac{3x}{6 \sin(x)} + \frac{1-\cos(x)}{3 \sin(x)}$

=$\frac{3x}{6 \sin(x)} + \frac{2\sin^2 (x/2)}{3 \sin(x)}$ = 1/2

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  • $\begingroup$ The first equation is correct, but the second should probably be a limit. A bit more explanation would be a good thing. $\endgroup$ – robjohn♦ Oct 17 '14 at 21:06

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