Find $\lim_{x\to0} \frac{\log(1+3x)}{f(x)}$ given $\lim_{x\to0} \frac{f(x)}{\sin(x)} = 2$

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).

• 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? – dan_fulea Feb 19 at 18:32
• Nitpick. If $\lim \frac {f(x)}{\sin x}$, I don't think we can say $f$ is "arbitrary". – fleablood Feb 19 at 19:16

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?

• 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. – Ren Eh Daycart Feb 19 at 22:39
• @RenEhDaycart $\lim_{x\to 0} \frac{\ln(1+3x)}{\sin x} = 3$, not $\frac{3}{2}$. – LHF Feb 19 at 22:41

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 $$\frac{\log(1+3x)}{f(x)}\sim_0 \frac{3x}{2x}=\frac 32.$$

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

• 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 $\lim_{x\to 0}\frac {\frac {\log(1+3x)}{\sin x}}{ \frac{f(x)}{\sin x}}$? – Ren Eh Daycart Feb 19 at 20:15
• 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 – Ren Eh Daycart Feb 19 at 20:27
• 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. – fleablood Feb 19 at 21:46
• 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. – Ren Eh Daycart Feb 19 at 22:37
• 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 $\lim \frac {\log(1+3x)}{f(x)}=\lim\frac {\log(1+3x)}{\sin x}\frac {\sin x}{f(x)} = 3*\frac 27 = \frac {6}{7}$ – fleablood Feb 19 at 22:50

Hint:

If the limits exist,

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

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