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I've been attempting to prove this limit (again). It must be a proof with epsilon delta definition.

$\frac{\sin(\log(x))}{\log(x)} \rightarrow 1$ when $x \rightarrow 1$

so far I have, given $\epsilon >0 $

$\lvert x-1\rvert < \delta$

$\Big \rvert \dfrac{\sin(\log(x))}{\log(x)} -1 \Big \lvert$ < $\Big \rvert \dfrac{\lvert \log(x)\rvert}{\log(x)} -1 \Big \lvert$ =$ \lvert 0 \rvert$

I've used $\lvert \sin(x) \rvert < \lvert x \rvert, x>0$

But this just seems wrong and i would like some help. Thanks in advance.

lingku
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  • Are you allowed to use that $\frac{\sin x}x\to1$ when $x\to0$? Without that it'll probably be difficult – Milten Feb 22 '24 at 14:14
  • How would you go, if that was the case? – lingku Feb 22 '24 at 14:17
  • And yes, what you have is wrong. You are saying, if $|y|<1$ then $|y-1| \le 0$, so $y=1$. Clearly that's not true (since there are many other possible values for $y$). In fact $|y-1| = 1-y \in [0,2]$ – Milten Feb 22 '24 at 14:19
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    It depends on the definition of $\sin$ and $\log$ you are working with. Are you allowed to use the fact that they are continuous functions? If so, by composition of two continuous functions is a continuous function, you are just doing $\lim_{y\to 0} \frac{\sin y}{y},$ which as Milten says, equals $1.$ – Adam Rubinson Feb 22 '24 at 14:22
  • The exercise should be done explicitly as an epsilon delta proof. – lingku Feb 22 '24 at 14:27
  • But what properties can you use? Continuity of the involved functions, known limits...? – Julio Puerta Feb 22 '24 at 14:34
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    As in this post, you can use the fact that for $x > 0$, $$ \cos(x) \leq \frac{\sin(x)}{x} \leq 1 $$ – Ben Grossmann Feb 22 '24 at 14:37
  • So far we have only covered inequalities for $\log(x)$, such af $\log(x)\geq \frac{x-1}{x}$ for x larger than 0 and variants of this, but everytime i've tried to utilize this and the upper bound for sin(x), i've failed and gotten $<\frac{1}{\lvert x-1 \rvert}$ when manipulating the function – lingku Feb 22 '24 at 14:38
  • All you need to know about $\log x$ here is that for some neighborhood about $x = 1, \log x \ne 0$ except at $x = 1$ and that $\lim_{x\to 1} \log x = 0$. Any inequalities that allow you to show these facts would be sufficient for the needs of this proof. – Paul Sinclair Feb 23 '24 at 18:17

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