If $\sin 0= 0$ and $2\cdot 0=0$, how can the answer be $1/2$?
Am I missing a step?
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3What is $\lim_{x\to0}\frac{\sin x}{x}$? – user84413 Aug 12 '14 at 20:24
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The most important reason for introducing the concept of limits in the first place is to deal with limits of expressions in which the numerator and denominator both approach $0$. That is what happens in the expression $\displaystyle f'(x)=\lim_{h\to0}\frac{f(x+h)−f(x)}h$. So it would be a serious mistake to think that if the numerator and denominator both have $0$ as their limits, then the limit of the fraction does not exist. – Michael Hardy Aug 12 '14 at 23:01
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Well you need to evaluate $$\lim_{x\to 0} \frac{\sin x}{2x}$$ We have \begin{align} &\lim_{x\to 0} \frac{\sin x}{2x}\\ &=\frac{1}{2} \left( \lim_{x \to 0} \frac{\sin x}{x} \right) \tag{*}\\ &=\frac{1}{2} \cdot 1\\ &=\frac{1}{2} \end{align}
To evaluate the limit inside the parenthesis in (*) we are using the well known theorem from Calculus, that $\lim_{x\to 0} \frac{\sin x}{x}=1$. For a nice geometric proof of this fact you can see Robjohn's answer here. If the OP is taking a calculus I class in the US it is very likely that your textbook uses a similar technique to prove this theorem. This is one of the so called fundamental limits that one has to know in Calculus I.
Of course once that theorem is proved it can be used freely in evaluating such limits. The point of these kinds of problems is precisely to get you to use that theorem. I hope this helps.
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$$\lim_{x \to 0} \frac{\sin x }{2x}=\frac{1}{2} \underbrace{\lim_{x \to 0} \frac{\sin x}{x}}_{1}=\frac{1}{2}$$
$\lim_{x \to 0} \frac{\sin x}{x} \overset{ l'Hospital}{=} \lim_{x \to 0} \cos x=1$
evinda
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4Using L'hopital's rule to evaluate this limit would be circular reasoning. Because if one uses L'hopital's rule then we need the derivative of $\sin x$, but to find it you need to evaluate a limit of the form $\lim_{x\to 0} \frac{\sin x}{x}$ – Cousin Aug 12 '14 at 20:27
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2To complement @JackDawkins's comment, it's circular if you're using the geometric definition of $\sin$ and it's likely that this is the definition being used. – Git Gud Aug 12 '14 at 20:28
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1@evinda whoops, the comment was meant for the OP. Put the comment in the wrong section. Apologies. – user137481 Aug 12 '14 at 20:29
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DeL'Hospital? It should be either l'Hôpital (modern French), or l'Hospital (old French). And spelled like L'Hopital's rule, it has 2 spelling mistakes. – Klaas van Aarsen Aug 12 '14 at 21:50
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Remember, we're not evaluating $\frac{\sin(x)}{2x}$ when $x=0$. If we were to do so, the answer would be undefined (since division by 0 is undefined). Instead, we are asking what is the limit of the value $\frac{\sin(x)}{2x}$ as $x$ gets closer and closer to 0. And that's a very different question.
This value can be determined using L'Hopital's rule.
NicNic8
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We have not been taught that in my class. Is there a simpler explanation? – user137452 Aug 12 '14 at 20:31
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@user137452 "simple" is a relative term. Try this website:http://www.larsoncalculus.com/calc10/content/proof-videos/chapter-1/section-3/proof-two-special-trigonometric-limits/#content-top – user137481 Aug 13 '14 at 00:18
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I guess you could use the definition of a limit, but I'm not sure it would be simple to do so. – NicNic8 Aug 13 '14 at 03:41
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The most important reason for introducing the concept of limits in the first place is to deal with limits of expressions in which the numerator and denominator both approach $0$.
That is what happens in the expression $\displaystyle f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)} h$.
So it would be a serious mistake to think that if the numerator and denominator both have $0$ as their limits, then the limit of the fraction does not exist.
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$\sin x\sim x$ around $0$, then $$\lim_{x\to 0}\frac{\sin x}{2x}=\lim_{x\to 0}\frac{x}{2x}=\frac{1}{2}.$$
idm
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