To find limit of $\lim_{x\to 0}\frac {\cos(\sin x) - \cos x}{x^4} $. I differentiated it using L Hospital's rule. I got $$\frac{-\sin(\sin x)\cos x + \sin x}{4x^3}\text{.}$$ I divided and multiplied by $\sin x$. Since $\lim_{x\to 0}\frac{\sin x}{x} = 1$, thus I got $\frac{1-\cos x}{4x^2}$.On applying standard limits, I get answer $\frac18$. But correct answer is $\frac16$. Please help.
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1You might have to use l'Hopital's rule more than once. – MasterOfBinary Nov 26 '13 at 18:49
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But on doing so, it gets extremely complicated – user2369284 Nov 26 '13 at 18:51
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$\lim_{x\to 0}\frac{\sin x}{x}\neq 0$! It equals $1$. – Antoine Nov 26 '13 at 18:56
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Hint: Look that the $x^4$ at the denominator, it's asking for some Taylor guy... – TheVal Nov 26 '13 at 18:59
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Without L'Hôpital
The expansions $\sin(x) = x - \frac{x^3}{6} + O(x^5)$ and $\cos(u) = 1 - \frac{u^2}{2} + \frac{u^4}{24} + O(u^6)$ combine joyfully to give $$ \cos(\sin x) - \cos(x) = \left(1 - \frac{x^2}{2} + \frac{5x^4}{24}\right) - \left(1-\frac{x^2}{2} + \frac{x^4}{24}\right) + O(x^5) $$ so finally, $$ \lim_{x\to 0} \frac{\cos(\sin x) - \cos(x)}{x^4} = \frac{5}{24}-\frac{1}{24} = \frac{1}{6}. $$
Siméon
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Using Prosthaphaeresis Formulas,
$$\cos(\sin x)-\cos x=2\sin\frac{x-\sin x}2\sin\frac{x+\sin x}2$$
So, $$\frac{\cos(\sin x)-\cos x}{x^4}=2\frac{\sin\frac{x-\sin x}2}{\frac{x-\sin x}2}\frac{\sin\frac{x+\sin x}2}{\frac{x+\sin x}2}\cdot\frac{x-\sin x}{x^3}\cdot\frac{x+\sin x}x\cdot\frac14$$
We know, $\lim_{h\to0}\frac{\sin h}h=1$
$$\text{Apply L'Hospital's Rule on }\lim_{x\to0}\frac{x-\sin x}{x^3}$$
$$\text{and we get }\lim_{x\to0}\frac{x+\sin x}x=1+\lim_{x\to0}\frac{\sin x}x$$
lab bhattacharjee
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@user2369284, $x\to0\implies x\pm\sin x\to0\implies\frac{x\pm\sin x}2\to0$ – lab bhattacharjee Nov 26 '13 at 18:59
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In the numerator, I divided by sin x to get $\frac{sin(sinx)}{sinx}cosx + 1$ – user2369284 Nov 26 '13 at 19:12
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@user2369284, we can write $$\frac{-\sin(\sin x)\cos x + \sin x}{4x^3}=\frac{-\frac{\sin(\sin x)}{\sin x}\sin x\cos x + \sin x}{4x^3}=\frac{-\sin2x+2\sin x}{8x^3}$$ – lab bhattacharjee Nov 26 '13 at 19:22
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But as we are getting -sin 2x and 2sin x, why not take sin x common and apply Lt x-> 0 (sin x)/x = 0. – user2369284 Nov 26 '13 at 19:26
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I didn't know that these simple formulas had a magnificent technical name "Prosthaphaeresis Formulas." I am feeling enlightened! :) By the way this provides the simplest solution with one application of L'Hospital (+1) – Paramanand Singh Nov 26 '13 at 19:45
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@ParamanandSingh, here(http://mathworld.wolfram.com/topics/TrigonometricIdentities.html) we can avail many Trigonometric Identities by name. In India, they formulae are taught without such name. – lab bhattacharjee Nov 27 '13 at 04:07
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@ParamanandSingh, do you know how to avoid even one use of L'Hospital? (http://math.stackexchange.com/questions/387333/are-all-limits-solvable-without-lhospital-rule-or-series-expansion) – lab bhattacharjee Nov 27 '13 at 04:07
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sorry to say but there is no easy route to $(x - \sin x)/x^{3}$ without L'Hospital. Check http://math.stackexchange.com/questions/437926/a-limit-problem-lim-limits-x-to-0-fracx-sin-sin-x-sin2xx6 for a long route – Paramanand Singh Nov 27 '13 at 04:38
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We have that
$$\frac{\cos x-1+\frac{x^2}2}{x^4} \to \frac1{24} \tag 1$$
therefore
$$\frac {\cos(\sin x) - \cos x}{x^4}=\frac{\sin^4 x}{x^4}\frac {\cos(\sin x) - 1+\frac{\sin^2 x}2}{\sin^4 x}-\frac{\cos x-1+\frac{x^2}2}{x^4}+\frac12\frac{x^2-\sin^2 x}{x^4}$$
therefore the given limit is equivalent to
$$\frac12\frac{x^2-\sin^2 x}{x^4}=\frac12\frac{x+\sin x}{x}\frac{x-\sin x}{x^3}\to \frac12 \cdot 2 \cdot \frac16=\frac16$$
using that $\frac{x-\sin x}{x^3}\to \frac16$ as shown here.
Proof of $(1)$, using that $\cos x=1-2\sin^2 \left(\frac x 2\right)$
$$\frac{\cos x-1+\frac{x^2}2}{x^4}=\frac{\frac{x^2}2-2\sin^2 \left(\frac x 2\right)}{x^4}=\frac18\frac{\left(\frac x 2\right)^2-\sin^2 \left(\frac x 2\right)}{\left(\frac x 2\right)^4} \to \frac1{24}$$
by the same previous result.
user
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