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?

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!
