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

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?

  • $\begingroup$ 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

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, $$\lim_{x\to0}\dfrac{x-\sin x}{x^3}=\dfrac1{3!}$$

  • $\begingroup$ 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

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}


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

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


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:

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

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



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.