Could you guys give me at least a hint at:

$$\lim_{x\to 0}\frac{\sin(x^2 + \frac{1}{x}) - \sin(\frac{1}{x}))}{x}$$ ? I already tried expanding the $\sin(x^2 + \frac{1}{x})$ but got nothing. Also, changing variables would not work because when $x\to 0$, $\frac{1}{x}\to\infty$ and I'm not allowed yet to use limits to infinity.


Based on the answers, I did:

$$\lim_{x\to 0}\frac{\sin(x^2 + \frac{1}{x}) - \sin(\frac{1}{x}))}{x} = \lim_{x\to0}\frac{\color{Blue}{\sin(\frac{x^2}{2})} \cos(\frac{x^2 + \frac{2}{x}}{2})\frac{x}{2}}{\color{Blue}{\frac{x^2}{2}}} = \color{Blue}{1} \cdot \lim_{x\to0}\cos(\frac{x^2+\frac{2}{x}}{2})\cdot\frac{x}{2} = 0$$ becuse we have a bounded function $\cos(\frac{x^2+\frac{2}{x}}{2})$ and a function that goes to $0$: $\frac{x}{2}$

  • $\begingroup$ (neither l'hospital rule...) $\endgroup$ – Marter Js May 22 '14 at 22:24
  • $\begingroup$ Are you allowed to use derivatives whatsoever? $\endgroup$ – DanZimm May 22 '14 at 22:27
  • $\begingroup$ @DanZimm nope :c $\endgroup$ – Marter Js May 22 '14 at 22:27

We have


now since $$\cos(x^2)-1\sim_0-\frac{x^2}2$$ and $$\sin(x^2)\sim_0 x^2$$ and the fact that the $\sin$ and $\cos$ functions are bounded we see readily that the desired limit is $0$


Hint: Convert the numerator to product using the identity $$ \sin p - \sin q = 2 \sin \frac{p-q}{2} \cos\frac{p+q}{2}$$ I think this is the simplest approach.

  • $\begingroup$ I updated my answer. Is this right? Thank you so much. $\endgroup$ – Marter Js May 22 '14 at 22:58
  • $\begingroup$ @Marter Js According to the identity, in your answer, the second sin in the numerator must be cos ... $\endgroup$ – Fermat May 22 '14 at 23:02
  • $\begingroup$ I changed. Cosine is also bounded. Is this right now? $\endgroup$ – Marter Js May 22 '14 at 23:05
  • $\begingroup$ Yes, but you have $\cos(\frac{x^2+\frac{2}{x}}{2})$... $\endgroup$ – Fermat May 22 '14 at 23:10
  • 1
    $\begingroup$ that's right. Thanks :) $\endgroup$ – Marter Js May 22 '14 at 23:15

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