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What is the limit of: $$\lim_\limits{x\to0}{\tan{x}-\sin{x}\over x^3}$$

So what I tried is:

$$\lim_\limits{x\to0}{{\sin{x}\over\cos{x}}-\sin{x}\over x^3}=\lim_\limits{x\to0}{{\sin{x}}({1\over\cos x}-1)\over x\cdot x^2}$$

From here, using the rule $\lim_\limits{x\to0}{\sin{x}\over{x}}=1$ it remains to evaluate $$\lim_\limits{x\to0}{{1\over\cos{x}}-1\over x^2}.$$ I tried changing it a lot of ways but it always gets messier so I'm not sure what to apply here to complete the question.

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  • $\begingroup$ Do you know Taylor polynomial ? If yes, find Taylor expansion of degree 3 of $\tan(x)-\sin(x)$. If no, apply l'Hospital rule 3 times. $\endgroup$ – Surb Oct 21 '18 at 15:24
  • $\begingroup$ And for further options, see this question and the ones linked there: math.stackexchange.com/questions/157903/… $\endgroup$ – Hans Lundmark Oct 21 '18 at 15:25
  • $\begingroup$ Apply L'Hospital's rule $\endgroup$ – Anik Bhowmick Oct 21 '18 at 15:29
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From your last line, $${{1\over\cos{x}}-1\over x^2}={1-\cos^2(x)\over \cos(x)(1+\cos(x))x^2}={1\over \cos(x)(1+\cos(x))}\cdot \left(\frac{\sin(x)}{x}\right)^2.$$ Can you take it from here?

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$$\frac{\tan x-\sin x}{x^3}=\frac{\tan x}{x}\frac{1-\cos x}{x^2}=\frac{\tan x}{x}\frac{2\sin^2\frac{x}{2}}{x^2}=\frac{1}{2}\frac{\tan x}{x}\Bigg(\frac{\sin \frac{x}{2}}{\frac{x}{2}}\Bigg)^2\to\frac{1}{2}$$

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$$\lim_{x \to 0 } \frac{\sec x - 1}{x^2} = \lim_{x \to 0 } \frac{\sec x \tan x}{2x}= \lim_{x \to 0}\frac{\sec x}{2}= \frac12$$

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$$\frac{\tan x - \sin x}{x^3} = \frac{\sin x - \sin x\cos x}{x^3\cos x} = \frac{\sin x}x \cdot\frac{1-\cos x}{x^2}\cdot\frac 1{\cos x}\to 1\cdot \frac 12\cdot \frac 1{\cos 0} = \frac 12$$

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Observe that, $x\to 0$ then $\cos{x}\to 1$ so, $\big(\frac{1}{\cos{x}}-1\big)\to 0$ and also $x^2\to 0$. Hence we can use L'Hospital's rule. $$\lim_{x \to 0 } \frac{(1/\cos x) - 1}{x^2} = \lim_{x \to 0 } \frac{\sec x - 1}{x^2}=\lim_{x \to 0 } \frac{\sec x \tan x}{2x}= \lim_{x \to 0}\frac{\sec x}{2}= \frac12$$

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