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Using the fact that $\lim\limits_{x \to 0} \frac{\sin(x)}{x}=1$, please help me show that

$$\lim\limits_{x \to 0} \frac{\tan(x)-\sin(x)}{\sin^2(x)}=0.$$

Because I am not familar with L'Hôpital's rule and Taylor's Theorem, please avoid the use of either in your solution.

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  • $\begingroup$ You can use this technique. $\endgroup$ – Mhenni Benghorbal Jun 13 '13 at 0:21
  • $\begingroup$ To all that thought that, sorry I appeared to be rude in my original question, English is not my native language and sometimes I'm not able to express myself as I would want to. My intent here wasn't to get easy answers, just understand how to get the answer, this is not homework by the way. Thank you all, helped me a lot. $\endgroup$ – Thums Jun 13 '13 at 0:30
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If you multiply top and bottom by $\cot x$, you get:

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

Using the taylor series for sine and cosine, we have:

$$\lim_{x\to 0}\frac{1}{1-\frac{x^2}{2!}+\ldots}\frac{\frac{x^2}{2!}+\ldots}{x-\frac{x^3}{3!}+\ldots}=\lim_{x\to 0}\frac{1}{1-\frac{x^2}{2!}+\ldots}\frac{\frac{x}{2!}+\ldots}{1-\frac{x^2}{3!}+\ldots}=0$$

after canceling an $x$ and plugging in $0$.

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  • $\begingroup$ Thank you, the key here was multiplying by the $cot(X)$, thank you very much. $\endgroup$ – Thums Jun 13 '13 at 0:21
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    $\begingroup$ @LuanCristianThums Actually, you didn't need to start with multiplying by $\cot x$. Just express everything in the original problem in terms of sines and cosines and with a little algebra you get the same thing. Especially in trig, the Master's advice holds: "There are many roads to the top of the mountain, Grasshopper". $\endgroup$ – Rick Decker Jun 13 '13 at 0:49
  • $\begingroup$ I think one can prove L'Hopital's rule for analytic functions using basically this method $\endgroup$ – Chris Brooks Jun 13 '13 at 5:17
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You could also finish Jared's first expression by doing this $$ \begin{align} &\lim_{x\rightarrow0}\frac{1-\cos x}{\cos x\sin x}\cdot\frac{1+\cos x}{1+\cos x}\\ =&\lim_{x\rightarrow0}\frac{1-\cos^2 x}{\cos x\sin x(1+\cos x)}\\ =&\lim_{x\rightarrow0}\frac{\sin^2 x}{\cos x\sin x(1+\cos x)}\\ =&\lim_{x\rightarrow0}\frac{\sin x}{\cos x(1+\cos x)} \end{align} $$ The limit of the denominator is 2 and the limit of the numerator is 0.

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  • $\begingroup$ That's what I used, thank you very much friend! $\endgroup$ – Thums Jun 13 '13 at 0:24
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$$\lim\limits_{x \to 0} \frac{\tan(x)-\sin(x)}{\sin^2(x)}=\lim\limits_{x \to 0} \frac{\frac{\sin(x)}{\cos (x)}-\sin(x)}{\sin^2(x)}=\lim\limits_{x \to 0} \frac{\frac{1}{\cos (x)}-1}{\sin(x)}=\lim\limits_{x \to 0} \frac{2\sin^2(\frac x2)}{2\sin(2x)}=.\lim\limits_{x \to 0} \frac x8=0$$

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  • $\begingroup$ I think I'm kind of lost here because trig identities are not my best, couldn't understand the whole path you took, but was able to get to the answer with the help of all of you, thanks! $\endgroup$ – Thums Jun 13 '13 at 0:26
  • $\begingroup$ @LuanCristianThums: You are welcome. $\endgroup$ – Paul Jun 13 '13 at 0:47
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The answer is zero, because the numerator is an odd function whose linear terms cancel, so it behaves as $a x^3$ as $x \to 0$, for some $a$, while the denominator behaves as $x^2$ in that limit. The latter point should be clear from your statement that $\sin{x}/x$ approaches $1$ as $x \to 0$. The former point may be reasoned as follows:

$$\sin{x} \sim x + \text{higher-order terms}$$ $$\tan{x} \sim x+ \text{higher-order terms}$$

$\tan{x}-\sin{x}$ is odd. Therefore there are no even terms in an expansion. Therefore, the lowest-order term is $a x^3$ for some $a$.

ADDENDUM

You can also use simple trig identities, e.g.,

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

and clearly, the limiting value of the RHS as $x \to 0$ is zero.

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  • $\begingroup$ Sorry mate, I don't get it. Anyway, my teacher want us to show the whole resolution and how we did get to the answer. $\endgroup$ – Thums Jun 12 '13 at 23:50
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    $\begingroup$ @LuanCristianThums: then this problem needs to be labeled as homework. And I hope nobody does your homework for you. $\endgroup$ – Ron Gordon Jun 12 '13 at 23:51
  • $\begingroup$ It's not homework, I'm studying for my test and that's how my teacher want it (on the test). I'm just here because I tried all I know without succeeding. What I meant to say is that I can't just write your statement. $\endgroup$ – Thums Jun 12 '13 at 23:53
  • $\begingroup$ @LuanCristianThums: what tools are known to you? $\endgroup$ – Ron Gordon Jun 12 '13 at 23:59
  • $\begingroup$ Variable substitution, multiplication by it's conjugate, writing $tan(x)$ like $\frac{sin(x)}{cos(x)}$, for example, mostly everything of this in order to get to the fundamental trig. limit identity and leave the indeterminate form. $\endgroup$ – Thums Jun 13 '13 at 0:03
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Do you know the derivatives of trig functions? If so: first, write $\tan x = \sin x/\cos x$. Then simplify the resulting fraction and multiply by $1 = \frac{x - 0}{x - 0}$, and use the limit you already know and a particular interpretation of what remains to see that the limit is indeed $0$. To start you off: \begin{align*} \lim_{x\to 0}{\frac{\tan(x)-\sin(x)}{\sin^2x}} &= \lim_{x\to 0}{\frac{\sin(x)/\cos(x)-\sin(x)}{\sin^2x}}\\ &= \lim_{x\to 0}{\frac{\sin(x)-\cos(x)\sin(x)}{\cos(x)\sin^2x}}\\ &= \lim_{x\to 0}\frac{1}{\cos x}\cdot\lim_{x\to 0}{\frac{1-\cos(x)}{\sin(x)}}\\ &= \lim_{x\to 0}{\frac{1-\cos(x)}{\sin(x)}}. \end{align*} Can you take it from here?

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  • $\begingroup$ Thank you @Stahl, I made it, using the multiplication by it's conjugate after multiplicating top and bottom of the original fraction by $cot(x)$. $\endgroup$ – Thums Jun 13 '13 at 0:23
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Hint: If $\cos(x)\ne1$, $$ \begin{align} \frac{\tan(x)-\sin(x)}{\sin^2(x)} &=\frac{\tan(x)(1-\cos(x))}{1-\cos^2(x)}\\ &=\frac{\tan(x)}{1+\cos(x)}\\ &=\frac{\sin(x)}{\cos(x)+\cos^2(x)}\\ &\to\frac02 \end{align} $$

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  • $\begingroup$ I need to read the previous answers more closely. Ron Gordon's and Rick Decker's answers have essentially the same identity. However, now I see that Ron Gordon did not add this part of his answer until I had started this one. So I will undelete it. $\endgroup$ – robjohn Jun 13 '13 at 0:15

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