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I can't figure out the next equation. The answer is negative infinity, but i don't know how to get there by using L'Hospital. The equation is:

$$\lim_{x\to 0} \frac{x-\sin(2x)}{x^3}$$

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    $\begingroup$ Which problem do you run into when you try to use L'Hospital's rule? $\endgroup$ – José Carlos Santos Jun 3 '19 at 9:45
  • $\begingroup$ If i get the derivative of the function i get: $$ -((2x+2xcos(2x)-3sin(2x))/x^4 $$ $\endgroup$ – Wouter Lommerse Jun 3 '19 at 9:48
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    $\begingroup$ Actually, you are supposed to get the derivative of the numerator and the derivative of the denominator, not the derivative of the quotient. $\endgroup$ – José Carlos Santos Jun 3 '19 at 9:50
  • $\begingroup$ Router: Have a look at examples en.m.wikipedia.org/wiki/L%27Hôpital%27s_rule $\endgroup$ – Peter Szilas Jun 3 '19 at 9:57
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A possible way is to use the following standard limit (which can also been easily shown using L'Hospital):

  • $\frac{1-\cos x}{x^2}\stackrel{x\to 0}{\longrightarrow}=\frac{1}{2}$

\begin{eqnarray*} \lim_{x\to 0} \frac{x-\sin(2x)}{x^3} & \stackrel{L'Hosp.}{\sim} & \frac{1-2\cos(2x)}{3x^2} \\ & = & \underbrace{4\frac{1-\cos(2x)}{3(2x)^2}}_{\stackrel{x\to 0}{\longrightarrow}\frac{4}{3}\cdot \frac{1}{2}} -\underbrace{\frac{\cos(2x)}{3x^2}}_{\stackrel{x\to 0}{\longrightarrow}\infty}\\ &\stackrel{x\to 0}{\longrightarrow}& -\infty \end{eqnarray*}

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Using $\sin x $$x-x^3/6$, it is simple to see that the limit diverges $$\lim_{x \rightarrow 0}=\frac{x-\sin 2x}{x^3}=\lim_{x \rightarrow 0} \frac{x-2x+8x^3/6}{x^3}=\lim_{x \rightarrow 0} \frac{-1}{x}+\frac{4}{3}=-\infty$$

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To apply L'hopital, it doesn't seem necessary to apply any identities to the trig part

$$\frac{x - \sin(2x)}{x^3}$$

differentiate top and bottom (independently)

$$\frac{1 - 2\cos(2x)}{3x^2}$$

then clearly as x tends to zero, the numerator becomes negative 1, nut the denominator becomes zero - giving you your result of a tending to negative infinity

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  • $\begingroup$ You cannot say the 'numerator becomes $-1$', since $\cos(2x)$ is an oscillating function, so it does not converge to anything at negative infinity. $\endgroup$ – Toby Mak Jun 3 '19 at 11:26
  • $\begingroup$ in his question, he has limit as x tends to zero, not infinity - if it was infinity, the squeeze or sandwich rule would work great $\endgroup$ – Cato Jun 3 '19 at 11:41
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    $\begingroup$ Sorry for the mistake, I couldn't see the $0$ from the computer screen :) $\endgroup$ – Toby Mak Jun 3 '19 at 11:53
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\begin{align*} \lim_{x \to 0}\frac{x-\sin 2x}{x^3}&=\lim_{x \to 0}\frac{x-2\sin x\cos x}{x^3}\\ &=\lim_{x \to 0}\frac{x-\sin x+\sin x-2\sin x\cos x}{x^3}\\ &=\lim_{x \to 0}\frac{x-\sin x}{x^3}+\lim_{x \to 0}\frac{\sin x(1-2\cos x)}{x^3}\\ &=\lim_{x \to 0}\frac{1-\cos x}{3x^2}+\lim_{x \to 0}\frac{1-2\cos x}{x^2}\\ &=\frac{1}{6}+\frac{-1}{0^2}\\ &=\frac{1}{6}+(-\infty)\\ &=-\infty. \end{align*}

Can you understand me?

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