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I've been asked to evaluate the limit using L'hospital's rule, or any simpler more elementary method.

$$\lim\limits_{x \to 0} {{x-\sin(x)}\over{x-\tan(x)}}$$

Applying L'Hospitals rule, I get: $${1-\cos(x)}\over{1-\sec^2(x)}$$

This still yields an indeterminate form, and applying L'Hospitals rule any more times makes things get messy fast. I think I'm stuck.

Other things I have tried include writing in terms of sine and cosine, both before and after applying the rule. I can't seem to wrestle it out of an indeterminate form.

How can I proceed?

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  • $\begingroup$ Try applying l'hospital a second time. And divide top and bottom by x afterwards. $\endgroup$ – randomgirl Nov 6 '15 at 17:37
  • $\begingroup$ alternatively why not expand $tan$ and $sin$ up to third order around $0$ and see what happens.. $\endgroup$ – tired Nov 6 '15 at 17:42
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We have the following:

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

which leads to

$$\lim_{x\to 0}\frac{-\cos^{2}x}{1+\cos(x)}=\frac{-1}{2}$$

which seems to be correct, according to WolframAlpha.

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  • $\begingroup$ D'oh! I tried this method, but I guess I lost a negative sign somewhere, because I ended up with 1/2, which was wrong. $\endgroup$ – Bassinator Nov 6 '15 at 17:40
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    $\begingroup$ You were close. Unlike the other answers, I'd recommand to try not to use L'Hospital's rule when it is not necessary because it often involves a lot of calculus. $\endgroup$ – MoebiusCorzer Nov 6 '15 at 17:41
  • $\begingroup$ @MoebiusCorzer that is an important point, (+1) $\endgroup$ – tired Nov 6 '15 at 17:43
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Yes apply L'Hopital's rule one more time!

$$\lim\limits_{x \to 0} {{x-\sin(x)}\over{x-\tan(x)}}= \lim\limits_{x \to 0}{{1-\cos(x)}\over{1-\sec^2(x)}}=\lim\limits_{x \to 0}{{\sin(x)}\over{-2\tan(x)\sec^2(x)}}=\lim\limits_{x \to 0}{{1}\over{-2\sec^3(x)}}= \frac{-1}{2}$$

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L'Hôpital's gets a little messy, but not too bad:

$$ \frac{x-\sin x}{x-\tan x} \to \frac{1-\cos x}{1-\sec^2 x} \to \frac{\sin x}{-2\sec^2 x\tan x} \to \frac{\cos x}{-2(\sec^4 x+2\sec^2x\tan^2 x)} $$

which equals $-\frac{1}{2}$ at $x = 0$.

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  • $\begingroup$ You can, of course, short-circuit that last one as in Alex's solution. $\endgroup$ – Brian Tung Nov 6 '15 at 17:41
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$$\lim\limits_{x \to 0} {{x-\sin(x)}\over{x-\tan(x)}}$$ Applying L'Hospitals rule, we get: $$\lim_{x\to 0}\frac{1-\cos(x)}{1-\sec^2(x)}$$ Applying L'Hospitals rule a second time, we get: $$\lim_{x\to 0}\frac{\sin(x)}{-2\sec^2(x)\tan(x)}$$ Applying L'Hospitals rule a third time, we get: $$\lim_{x\to 0}\frac{\cos(x)}{-2\sec^4(x)-4\sec^2(x)\tan^2(x)}$$ $$=-\frac{1}{2}$$

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