<s>[INST] When not to use $ \lim\limits_{x\rightarrow 0}\frac{\log\left( 1+x\right) }{x}=1 $ My Teacher has told to use this as a standard result so when question is in this I directly apply this result as in this
$$\lim\limits_{x\rightarrow 0}\dfrac{\log\left( 5+x\right) }{x}-\dfrac{\log\left( 5-x\right) }{x}=2/5$$
but here
$$\lim\limits_{x\rightarrow 0}\dfrac{\log\left( 1+2x\right) -2{\log\left( 1+x\right)} }{x^2}=0 $$
doesn't work because solution is give using expansion of $\log\left( 1+x\right)$
I am interested to know why we cant simply multiply and divide 2x to  log(1+2x) to convert it in known form like log(1+2x) / 2x *2x yielding 2x
So what is limitation of result when to use it and when to not?
 [/INST] The proposed result is not valid.
Having said that, you may be also interested in the power series method:
\begin{align*}
\ln(1 + x) = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \ldots
\end{align*}
whenever $|x| < 1$. If we restrict the values of $x$ to $(-1/2,1/2)$, we may claim as well that
\begin{align*}
\ln(1 + 2x) = 2x - 2x^{2} + \frac{8x^{3}}{3} - \ldots
\end{align*}
Therefore the proposed limit is given by
\begin{align*}
\lim_{x\to 0}\frac{\ln(1 + 2x) - 2\ln(1 + x)}{x^{2}} & = \lim_{x\to 0}\frac{2x - 2x^{2} - 2x  + x^{2} + o(x^{2})}{x^{2}} = \lim_{x\to 0}\frac{-x^{2} + o(x^{2})}{x^{2}} = -1
\end{align*}
Hopefully this helps!
</s>