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It's assumed that $\lim_{x\to0} \frac{f(x)}{\sin(x)} = 2$.

Find $$\lim_{x\to0} \frac{\log(1+3x)}{f(x)}$$

I don't think that it would work out by a random plugging.

Let $f(x) = 2\sin(x).$ $$\lim_{x\to0}\frac{\log(1+3x)}{f(x)} = \lim_{x\to0}\frac{\log(1+3x)}{2\sin(x)} = [\log(1+3x) = 3x + O(x^2)] = \lim_{x\to0}\frac{3x+O(x^2)}{2\sin(x)} = \frac{3}{2}.$$

But what did I miss? I cannot find a way to prove whether these two limits are connected (or that the result is unique).

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  • $\begingroup$ Note that you can write:$$\lim_{x\to0} \frac{\log(1+3x)}{f(x)}=\lim_{x\to0} \frac{\log(1+3x)}{\sin x} \cdot \frac{\sin x}{f(x)}\ .$$ Is it possible to split the R.H.S as a product of two limits? Why? $\endgroup$ – dan_fulea Feb 19 at 18:32
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    $\begingroup$ Nitpick. If $\lim \frac {f(x)}{\sin x}$, I don't think we can say $f$ is "arbitrary". $\endgroup$ – fleablood Feb 19 at 19:16
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How about using these (well-known) limits:

$$\lim_{x\to 0} \frac{\ln (1+x)}{x}=1,\ \ \lim_{x\to 0} \frac{\sin x}{x}=1$$

and writing:

$$\lim_{x\to 0}\frac{\ln(1+3x)}{f(x)} = 3\lim _{x\to 0} \left[\frac{\ln (1+3x)}{3x}\cdot \frac{x}{\sin x}\cdot \frac{\sin x}{f(x)}\right]$$

Can you end it from here?

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  • $\begingroup$ What if we happen to have $f(x) = \left(\frac{1}{2} \cdot sin(x) \right) + log(1+3x)$? Then the limit of $\frac{f(x)}{sin x}$ is $\frac{1}{2} + \frac{3}{2} = 2$, but the limit of $\frac{log(1+x)}{f(x)}$ seems to be $\frac{3}{\frac{1}{2} + 3}$ = 6/7. $\endgroup$ – Ren Eh Daycart Feb 19 at 22:39
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    $\begingroup$ @RenEhDaycart $\lim_{x\to 0} \frac{\ln(1+3x)}{\sin x} = 3$, not $\frac{3}{2}$. $\endgroup$ – LHF Feb 19 at 22:41
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With equivalence of functions near $0$:

The hypothesis means that $f(x)\sim_0 2\sin x$

On the other hand $\sin x\sim_0 x$ and $\ln(1+x)\sim_0 x$, so $$\frac{\log(1+3x)}{f(x)}\sim_0 \frac{3x}{2x}=\frac 32.$$

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When $f(x) \ne 0; \sin x \ne 0$ we have

$\frac {\log{1+3x}}{f(x)} = \frac {\log{1+3x}}{\sin x}\frac {\sin x}{f(x)}=\frac {\log{1+3x}}{\sin x}\cdot \frac 1{\frac {f(x)}{\sin x}}$

so if $\lim_{x\to 0}\frac {\log{1+3x}}{\sin x}$ exists than

$\lim_{x\to 0}\frac {\log{1+3x}}{f(x)} = \lim_{x\to 0}\frac {\log{1+3x}}{\sin x}\cdot\frac 1{\lim_{x\to 0}\frac {f(x)}{\sin x}} = \frac 12\lim_{x\to 0}\frac {\log{1+3x}}{\sin x}$

And we can use L'hopital to solve $\lim_{x\to 0}\frac {\log{(1+3x)}}{\sin x}=\lim_{x\to 0} \frac {3\frac 1{1+3x}}{\cos x}= \frac {3\frac 11}{1}=3$

So $\lim_{x\to 0}\frac {\log{1+3x}}{f(x)}=\frac 12\lim_{x\to 0}\frac {\log{1+3x}}{\sin x}=\frac 32$.

====== old answer =====

You don't let $f(x) = 2\sin x$ but you note that as $\lim_{x\to 0} \frac {f(x)}{\sin x} = 2$ then for any $\lim_{x\to 0} h(x) = K$ then $\lim_{x\to 0} h(x)\frac {f(x)}{\sin x} = 2K$.

So if $\lim_{x\to 0} \frac {\log(1 +3x)}{f(x)}$ exists

then $\lim_{x\to 0} \frac {\log{(1 +3x)}}{f(x)}= \frac12\lim_{x\to 0} \frac {\log{(1 +3x)}}{f(x)}\frac {f(x)}{\sin x}=\frac 12\lim_{x\to 0}\frac {\log{(1+3x)}}{\sin x}$

And we can use L'hopital to solve $\frac 12\lim_{x\to 0}\frac {\log{(1+3x)}}{\sin x}=\frac 12\lim_{x\to 0} \frac {3\frac 1{1+3x}}{\cos x}= \frac 12\frac {3\frac 11}{1}=\frac 32$.

I suppose I skirted the issue as to whether it is possible for $\lim_{x\to 0} \frac {\log(1 +3x)}{f(x)}$ to NOT exist.

But it must, as $\lim_{x\to 0} \frac {\log(1+3x)}{\sin x}=3$ and $\lim_{x\to 0}\frac {f(x)}{\sin x}=2$ do exist it must follow that $\frac 32 = \lim_{x\to 0}\frac {\frac {\log(1+3x)}{\sin x}}{ \frac{f(x)}{\sin x}}=\lim_{x\to 0}\frac {\log(1+3x)}{f(x)}$.

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  • $\begingroup$ Regarding the issue of whether it is possible for $\lim_{x\to 0} \frac {\log(1 +3x)}{f(x)}$ to NOT exist *** How do you know that we don't have x arbitrarily close to zero with $f(x)=0$ for those values of x? If there's no interval around zero with $f(x)$ non-zero for all values x in such an interval, then how can you use the expression $\lim_{x\to 0}\frac {\frac {\log(1+3x)}{\sin x}}{ \frac{f(x)}{\sin x}}$? $\endgroup$ – Ren Eh Daycart Feb 19 at 20:15
  • $\begingroup$ A specific example: what if we happen to have the following? $f(x) =log(1+3x)$, if $\exists n \in \mathbb{N}$ such that $x = \left(\frac{1}{\pi}\right)^n$, $f(x)=2sin(x)$, otherwise $\endgroup$ – Ren Eh Daycart Feb 19 at 20:27
  • $\begingroup$ If that were so then the limit of $\frac {f(x)}{\sin x}$ couldn't exist. "If there's no interval around zero with f(x) non-zero for all values x in such an interval" That can't be the case as bout $\frac {\log{1+3x}}{\sin x}$ and $\frac{f(x)}{\sin x} $ have limits. $\endgroup$ – fleablood Feb 19 at 21:46
  • $\begingroup$ Thanks! I have a better challenge: what if we happen to have $f(x) = \left(\frac{1}{2} \cdot sin(x) \right) + log(1+3x)$? Then the limit of $\frac{f(x)}{sin x}$ is $\frac{1}{2} + \frac{3}{2} = 2$, but the limit of $\frac{log(1+x)}{f(x)}$ seems to be $\frac{3}{\frac{1}{2} + 3}$ = 6/7. $\endgroup$ – Ren Eh Daycart Feb 19 at 22:37
  • $\begingroup$ No. $\lim \frac {\frac 12\sin (x) + \log(1+3x)}{\sin x}=\lim (\frac 12 +\frac {\log(1+3x)}{\sin x}) = \frac 12 +3 = \frac 72\ne 2$ and so $\lim \frac {\log(1+3x)}{f(x)}=\lim\frac {\log(1+3x)}{\sin x}\frac {\sin x}{f(x)} = 3*\frac 27 = \frac {6}{7}$ $\endgroup$ – fleablood Feb 19 at 22:50
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Hint:

If the limits exist,

$$\lim_{x\to0} \frac{f(x)}{\sin(x)}\cdot\lim_{x\to0} \frac{\log(1+3x)}{f(x)}=\lim_{x\to0} \frac{f(x)}{\sin(x)}\frac{\log(1+3x)}{f(x)}=\lim_{x\to0} \frac{\log(1+3x)}{\sin(x)}.$$

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  • $\begingroup$ I think you forgot a factor of $\frac{1}{2}$ in front of the last limit. $\endgroup$ – LHF Feb 19 at 18:43
  • $\begingroup$ @Atticus: no, where would it come from ? But I just noticed a typo, the second expression was the same as the first. $\endgroup$ – Yves Daoust Feb 19 at 21:03
  • $\begingroup$ The limit given as hypothesis is $2$, so it's inverse would be $\frac{1}{2}$. $\endgroup$ – LHF Feb 19 at 21:05
  • $\begingroup$ @Atticus: I know but read my identities again. $\endgroup$ – Yves Daoust Feb 19 at 21:06
  • $\begingroup$ Yes, sorry, I read it wrongly. $\endgroup$ – LHF Feb 19 at 21:08

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