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I don't know if I apply for this case sin (a-b), or if it is the case of another type of resolution, someone with some idea without using derivation or L'Hôpital's rule? Thank you.

$$\lim_{x\to0}\frac{\sin(x^2+\frac{1}{x})-\sin\frac{1}{x}}{x}$$

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  • $\begingroup$ What is $\sin(A+B)?$ $\endgroup$ – Will Jagy Jun 18 '13 at 20:46
  • $\begingroup$ I edited, sin (a + b) = sina(cosb) + senb(cosa). $\endgroup$ – user73276 Jun 18 '13 at 20:50
  • $\begingroup$ You will find the fact that $\lim_{x\to 0}\frac{\cos x^2-1}{x}=0$ handy. To show this, multiply top and bottom by $\cos x^2+1$. $\endgroup$ – André Nicolas Jun 18 '13 at 20:51
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Using the identity $$ \sin(A)-\sin(B)=2\sin\left(\frac{A-B}{2}\right)\cos\left(\frac{A+B}{2}\right) $$ we get $$ \begin{align} \lim_{x\to0}\frac{\sin\left(x^2+\frac1x\right)-\sin\left(\frac1x\right)}{x} &=\lim_{x\to0}\frac{2\sin\left(\frac{x^2}{2}\right)\cos\left(\frac{x^2}{2}+\frac1x\right)}{x}\\ &=\lim_{x\to0}x\frac{\sin\left(\frac{x^2}{2}\right)\cos\left(\frac{x^2}{2}+\frac1x\right)}{\frac{x^2}{2}}\\ &=\lim_{x\to0}x\cos\left(\frac{x^2}{2}+\frac1x\right)\lim_{x\to0}\frac{\sin\left(\frac{x^2}{2}\right)}{\frac{x^2}{2}}\\[12pt] &=0\cdot1 \end{align} $$

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We have $$\frac{\sin(x^2+\frac{1}{x})-\sin\frac{1}{x}}{x}=\frac{\sin x^2\cos\frac{1}{x}+\sin\frac{1}{x}\cos x^2-\sin\frac{1}{x}}{x}$$ and since $\sin x^2\sim_0 x^2$ and $\left|\cos \frac{1}{x}\right|\leq 1$ and since $\cos x^2\sim_0 1-\frac{x^4}{2}$ we can see easily that the given limit is $0$.

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  • $\begingroup$ This works because we can factor $$\sin(1/x)\cos(x^2)-\sin(1/x)=\sin(1/x)(\cos(x^2)-1)$$ $\endgroup$ – robjohn Jun 19 '13 at 0:30
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Using $\sin\left(a\right)-\sin\left(b\right)=2\sin\left(\frac{a-b}{2}\right)\cos\left(\frac{a+b}{2}\right)$ and $\cos\left(a+b\right)=\cos\left(a\right)\cos\left(b\right)-\sin\left(a\right)\sin\left(b\right)$: $$L=\lim_{x\rightarrow 0}\frac{2\sin\left(x^2/2\right)\cos\left(x^{-1}+x^2/2\right)}{x}\\=\lim_{x\rightarrow 0}\frac{2\sin\left(x^2/2\right)\left(\cos\left(x^2/2\right)\cos\left(x^{-1}\right)-\sin\left(x^2/2\right)\sin\left(x^{-1}\right)\right)}{x}\\=\lim_{x\rightarrow 0}\frac{\sin\left(x^2\right)\cos\left(x^{-1}\right)-2\sin^2\left(x^2/2\right)\sin\left(x^{-1}\right)}{x}\\=\lim_{x\rightarrow 0}x\cos\left(1/x\right)-\lim_{x\rightarrow 0}\frac{x^3\sin\left(1/x\right)}{2}=0.$$

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Perhaps not an elegant proof but I considered this.

Use a Taylor series for $\sin$.

$\sin(x)=x+ a_1x^3 + a_2x^5 + ...$

Let $A$ be the value of the limit.

Then we get $A=\dfrac{x^2+\frac{1}{x}-\frac{1}{x}+a_1(x^2+\frac{1}{x})^3-a_1(\frac{1}{x})^3+\ldots}{x} = \dfrac{x^2+a_1(x^2+\frac{1}{x})^3-a_1(\frac{1}{x})^3)+\ldots}{x}$.

This equals $A =\dfrac{x^2}{x}+\dfrac{a_1((x^3+1)^3-1)}{x^4}+\dfrac{a_2((x^3+1)^5-1)}{x^6}+\ldots$

Now use big $O$ notation to rewrite as follows : $A=\dfrac{O(x^2)}{x}+\dfrac{a_1O(x^9)}{x^4}+\dfrac{a_2O(x^{15})}{x^6}+\ldots$ hence $A=0 + a_1 0+a_20 + \ldots =0$. To justify that infinite sum notice that each limit $O(x^a)/x^b$ goes faster to $0$ than $x$ does.

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  • $\begingroup$ I tried to avoid trig formula's to show it can be done without. $\endgroup$ – mick Jun 18 '13 at 22:31
  • $\begingroup$ The big-O notation can easily be abused when used in a series. Here you have to show that the constants in the big-O estimates converge when summed. I don't think this can be easily done here since the cancellation in the tail of the series is important to keep sin and cos bounded especially near $0$ because of $\frac1x$. $\endgroup$ – robjohn Jun 19 '13 at 0:26
  • $\begingroup$ I disagree Rob. $a_n$ goes to $0$ so those are not a problem. Consider $\dfrac{O(x^9)}{x^4}$ if we rewrite that as $ \dfrac {~C x^9}{x^4}$ we also get limit $0$ for that term no matter how large $C$ is. $\endgroup$ – mick Jun 19 '13 at 22:13
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    $\begingroup$ Actually, as $x\to0$, it is the smaller powers of $x$ that dominate. Thus, $\frac{(x^3+1)^3-1}{x^4}=O(1/x)$ not $O(x^5)$ and $\frac{(x^3+1)^5-1}{x^6}=O(1/x^3)$ not $O(x^9)$. These blow up as $x\to0$. $\endgroup$ – robjohn Jun 20 '13 at 0:22
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    $\begingroup$ the difficulty with $\frac{(x^3+1)^3-1}{x^4}=O(1/x)$ is the real problem with your proof. My latter comment was to support my comment that you must have some knowledge of the coefficients related to the big-O estimates to guarantee convergence. For example, $$ \sum_{k=0}^\infty n!x^{4n+1}=O(x)+O(x^5)+O(x^9)+\dots $$ yet, the series converges only at $x=0$. $\endgroup$ – robjohn Jun 20 '13 at 21:59

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