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$$\lim_{x \to 0} \frac{\cos(\sin x)- \cos x}{x^4}$$

My approach

Using L'Hospital's rule we get $$ \lim_{x \to 0}\frac{\sin x - \sin(\sin x)\cos x}{4x^3} $$
Why can't we simplify this as,

$$ \begin{split} &\implies \lim_{x \to 0} \frac{\sin x}{x}\cdot\frac{1}{4x^2}-\frac{\sin(\sin x)\cdot(\sin x)}{(x\cdot\sin x)}\frac{\cos x}{4x^2}\text{ (dividing and multiplying sinx)} \\ &\implies \lim_{x \to 0}\frac{1-\cos x}{4x^2} \\ &=\frac 18 \end{split} $$ But the answer is $\frac 16$. Which I got by using L'Hospital's rule 2 more times.

Why is my simplification wrong?

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  • $\begingroup$ Your post has been formatted, but see this link for information on how to typset Maths in your future questions $\endgroup$ – lioness99a May 16 '17 at 10:31
  • $\begingroup$ This is a common mistake. In general you can not replace an expression by its limit while evaluating limit of a complicated expression. Such replacements are allowed only in very specific cases. See this answer math.stackexchange.com/a/1783818/72031 $\endgroup$ – Paramanand Singh May 16 '17 at 13:08
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If L'Hosiptal is not mandatory,

$$\dfrac{\cos(\sin x)-\cos(x)}{x^4}=\dfrac24\cdot\dfrac{\sin\dfrac{x-\sin x}2}{\dfrac{x-\sin x}2}\cdot\dfrac{\sin\dfrac{x+\sin x}2}{\dfrac{x+\sin x}2}\cdot\dfrac{x-\sin x}{x^3}\cdot\dfrac{x+\sin x}x$$

$$\dfrac{x+\sin x}x=1+\dfrac{\sin x}x$$

Using using Are all limits solvable without L'Hôpital Rule or Series Expansion, $$\lim_{x\to0}\dfrac{x-\sin x}{x^3}=\dfrac1{3!}$$

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  • $\begingroup$ Used mathworld.wolfram.com/ProsthaphaeresisFormulas.html $\endgroup$ – lab bhattacharjee May 16 '17 at 10:28
  • $\begingroup$ I know how to solve it without L'hospital rule, but my question was not how to solve it rather why is my method wrong. $\endgroup$ – TheLostGuardian0 May 16 '17 at 10:28
  • $\begingroup$ @TheLostGuardian0, How have you managed with $$\dfrac{\dfrac{\sin x}x}{x^2}$$ $\endgroup$ – lab bhattacharjee May 16 '17 at 10:30
  • $\begingroup$ I used lim f(x)*g(x)=lim f(x) * lim g(x) $\endgroup$ – TheLostGuardian0 May 16 '17 at 10:32
  • $\begingroup$ And so I made it (Sinx/x)*(1/x^2) $\endgroup$ – TheLostGuardian0 May 16 '17 at 10:37
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You can use substitution to solve. Let $t=\sin x$ and then $x=\arcsin t$. So \begin{eqnarray} &&\lim_{x \to 0} \frac{\cos(\sin x)- \cos x}{x^4}\\ &=&\lim_{t \to 0} \frac{\cos t- \sqrt{1-t^2}}{(\arcsin t)^4}\\ &=&\lim_{t \to 0} \frac{\cos t- \sqrt{1-t^2}}{t^4}\frac{t^4}{(\arcsin t)^4}\\ &=&\lim_{t \to 0} \frac{(1-\frac{1}{2}t^2+\frac{1}{4!}t^4+O(t^6))-(1-\frac12t^2-\frac{1}{8}t^4+O(t^6))}{t^4}\\ &=&\frac16. \end{eqnarray}

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Doing l'Hospital right away here can be a huge pain in a very sensitive zone. I'd propose first some power series:

$$\frac{\cos\sin x-\cos x}{x^4}=\frac1{x^4}\left(1-\frac{\sin^2x}{2!}+\frac{\sin^4x}{4!}-\ldots-\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\ldots\right)\right)=$$$${}$$

$$=\frac{-\frac{\sin^x}2+\frac{\sin^4x}{24}-\ldots+\frac{x^2}2-\frac{x^4}{24}-\ldots}{x^4}=\frac{-12\sin^2x+\sin^4x+12x^2-x^4+\ldots}{24x^4}$$

Observe that "the dots..." above are terms for which either $\;\sin x\;$ or $\;x\;$ are raised to a power higher than $\;4\;$ and thus they tend to zero when divided over $\;x^4\;$ , so we're left only with the above and now we apply l'Hospital:

$$\lim_{x\to0}\frac{-12\sin2x+4\sin^3x\cos x+24x-4x^3}{96x^3}=\lim_{x\to0}\frac{-24\cos2x+3\sin^22x-4\sin^4x+24-12x^2}{288x^2}=$$

$$=\lim_{x\to0}\frac{48\sin2x+6\sin4x-16\sin^3x\cos x-24x}{576x}=$$

$$=\lim_{x\to0}\frac{96\cos2x+24\cos4x-12\sin^22x+16\sin^4x-24}{576}=\frac{96+24-24}{576}=\frac16$$

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