Prove that $\lim_{x \to 0} \frac {f(3x)}{\ln(1+4x)} = \frac94$ if $\lim_{x \to 0} \frac {f(x)}{x} = 3$

Question

I need to prove the following: $f(x)$ defined on a neighborhood of $x=0$ such that: $$ \lim_{x \to 0} \frac {f(x)}{x} = 3 $$ I need to prove that: $$ \lim_{x \to 0} \frac {f(3x)}{\ln(1+4x)} = 2.25 $$ Basically, what I did is: $$ \lim_{x \to 0} \frac {f(3x)}{\ln(1+4x)} = \lim_{x \to 0} \frac {f(3x)}{x} \cdot \frac{x}{\ln(1+4x)} = \lim_{x \to 0} \frac {f(3x)}{x} \cdot \frac{4x}{\ln(1+4x)} \cdot \frac{1}{4} $$ Now I'm stuck here with 2 questions:

1) Can I say that since $x \to 0$ it is hold that: $$ \lim_{x \to 0} \frac{f(x)}{x} = \lim_{x \to 0} \frac{f(3x)}{x} $$

If so, then I can simply switch this expression to $3$.

2) The other question is about the $\ln$ expression, I know the identity that: $$ \lim_{x \to 0} \frac {\ln(1+x)}{x} = 1 $$

But in my expression I need to switch the denominator with the numerator in order to use this Identity. How can I do so?

I'm not sure about any of my steps so please correct me where I'm wrong and don't just offer your own solution.

Thank you

Answer 1

For the first question note that $y := 3x \to 0$ for $x \to 0$, but you have to replace $x$ by $3x$ everywhere, so $$ \frac{f(3x)}{x} = \frac{f(3x)}{3x} \cdot 3 = \frac{f(y)}{y} \cdot 3 \to 3 \cdot 3 = 9 $$ For the second problem, note that (with $z := 4x$ we have $z \to 0$ for $x \to 0$) it holds: $$ \frac{4x}{\log(1 + 4x)} = \frac 1{\log(1+z)/z} \to \frac 11 = 1 $$ So we have $$ \frac{f(3x)}{x} \cdot \frac{4x}{\log(1 + 4x)} \cdot \frac 14 \to 9 \cdot 1 \cdot \frac 14 = \frac 94 $$

Answer 2

Hint: Why not write the limit as $$\lim_{x \to 0} \frac {f(3x)}{3x}\times \frac {4x}{\ln (1+4x)} \times \frac {3}{4}??$$

Answer 3

Another observation is : $$\lim_{x \to 0 }\frac{f(x)}{x}=3 \to f(x) \sim 3x+o(x^2)\\ \lim_{x \to 0 }\frac{f(3x)}{\ln(1+4x)}=\lim_{x \to 0 }\frac{3(3x)+o(x^3)}{\ln(1+4x)}=\\\lim_{x \to 0 }\frac{9x+o(x^2)}{4x+o(x^2)}=\lim_{x \to 0 }\frac{9x}{4x}=\frac{9}{4}=2.25$$ in your way :take $3x=a \to x=\frac{a}{3} \to 0$ $$\lim_{x \to 0 }\frac{f(3x)}{ln(1+4x)}=\\ \lim_{a \to 0 }\frac{f(a)}{ln(1+4\frac{a}{3})}=\\ \lim_{a \to 0 }\frac{f(a)}{a}\frac{a}{ln(1+4\frac{a}{3})}=\\ 3.\lim_{a \to 0 }\frac{a}{ln(1+4\frac{a}{3})}=\\ 3.\lim_{a \to 0 }\frac{a}{4\frac{a}{3}}=2.25$$

Answer 4

To prove this sort of limit, you will need to go back to the $\delta-\epsilon$ definition but first, it is helpful to derive, in a correct but non-rigorous manner, what the answer will be, because you can then use those steps to determine what $\delta$ must be in terms of $\epsilon$.

Your first step trick is good: $$ \lim_{x\to 0} \frac{f(3x)}{\ln(1+4x)} = \lim_{x\to 0} \left(\frac{f(3x)}{3x}\right) \left( \frac{3x}{\ln(1+4x)} \right) = \left(\lim_{x\to 0} \frac{f(3x)}{3x}\right) \left( \lim_{x\to 0} \frac{3x}{\ln(1+4x)} \right) $$ where the second step in that equation is only justified if each individual limit exists -- that is what I mean by a non-rigorous manner.

Then you can find those limits by changing variables to $y=3x$ and $u= \ln(1+4x)$ so that in the first limit, $x=y/3$ and in the second, $x=\frac{e^u-1}{4}$. $$ \left(\lim_{x\to 0} \frac{f(3x)}{3x}\right) \left( \lim_{x\to 0} \frac{3x}{\ln(1+4x)} \right) =\left(\lim_{y\to 0} \frac{f(y)}{y}\right) \left( \lim_{u\to 0} \frac34 \frac{e^u-1}{u} \right)= \frac94 \lim_{u\to 0} \frac{e^u-1}{u} $$