Understanding that both $\lim_{y\to 0} \frac{y}{ln(y+1)}$ and $\lim_{y\to 0} \frac{\ln(y+1)}{y}$ is equal to $1$. Question:
I want to understand that both: 
$\lim_{y\to 0} \frac{y}{ln(y+1)}$ and $\lim_{y\to 0} \frac{\ln(y+1)}{y}$
is equal to $1$ using algebraic manipulations and standard limits (not L'Hospitals rule or series expansions).
Reasoning:
First of all is seems logical that when we switch the numerator and denominator that the limit still would converge to the same value. But is this always true?
In this particular case i tried to prove it but my result were the following:
$\lim_{y\to 0} \frac{\ln(y+1)}{y} = \lim_{y\to 0} \frac{1}{y}\ln(y+1) = \ln(y+1)^{\frac{1}{y}} = \ln(e) = 1$
$\lim_{y\to 0} \frac{y}{\ln(y+1)} = \lim_{y\to 0} y(\ln(y+1))^{-1} =$ ??? $= \ln(e) = 1$

It is true that when $\displaystyle\lim_{x\to x_0} f(x) = \ell$ and $\ell \neq 0$, then $\displaystyle \lim_{x\to x_0} \frac{1}{f(x)}= \frac{1}{\ell}$.
Proof:
Let $g(x) = \frac{1}{f(x)}$. Then, $\displaystyle \lim_{x\to x_0} g(x) = \lim_{x\to x_0} \frac{1}{f(x)} \stackrel{*}{=} \frac{\displaystyle\lim_{x\to x_0} 1}{\displaystyle\lim_{x\to x_0}f(x)}= \frac{1}{\ell}.$
Here, since $\displaystyle \lim_{y \to 0} \frac{\ln(y+1)}{y}=1 \implies \lim_{y\to 0} \frac{y}{\ln(y+1)} = \frac{1}{1} = 1.$

$*$ Generally, when $\displaystyle \lim_{x\to x_0} f(x) = a$ and $\displaystyle\lim_{x\to x_0} g(x) = b\neq 0,$ then $\displaystyle \lim_{x\to x_0} \frac{f(x)}{g(x)}$ exists and equals to $\displaystyle \frac{\lim_{x\to x_0} f(x)}{\lim_{x\to x_0} g(x)}=\frac ab. $
