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?
 A: In the first case, you have the limit
$$\lim_{x\to 0}\frac{\ln(5+x)-\ln(5-x)}{x}.$$
Since $\lim\limits_{x\to 0}(\ln(5+x)-\ln(5-x))=0$ and $\lim\limits_{x\to 0}(x)=0$, L'Hopital's rule says that the above limit is equal to
$$\lim_{x\to 0}\left(\frac{1}{x+5}+\frac{1}{5-x}\right)=\frac25.$$
(I wouldn't call this a direct application of $\lim\limits_{x\to 0}\frac{\ln(1+x)}{x}=1$, but it's possible there's another solution using that fact.)
For the second problem, we again have the so-called indeterminate form $\frac00$, so applying L'Hospital's rule gives
$$\lim_{x\to 0}\frac{\frac{2}{1+2x}-\frac{2}{1+x}}{2x}=\lim_{x\to 0}\frac{\frac{-2x}{(1+2x)(1+x)}}{2x}=\lim_{x\to 0}\frac{-1}{(1+2x)(1+x)}=-1.$$
A: Though L'Hospitals rule as used by @pancini is a better alternative to your problem, I would just show that the aforementioned 'result' can be used on both the problems.
First problem:
Rewrite as
$$\lim_{x\to 0} \frac{\ln\frac{5+x}{5-x}}{x}$$
The numerator is $$\ln(1+ \frac{2x}{5-x})$$
Hence the limit becomes
$$\lim_{x \to 0} \frac{\ln(1+ \frac{2x}{5-x})}{(\frac{2x}{5-x})} × \frac{2}{5-x} $$
Here I use your result ,but take the expression$$\frac{2x}{5-x}$$which also tends to zero,hence your "result" is applicable modifying it as,$$ \lim \frac{2}{5-x} = \frac{2}{5} $$
In the second problem you proceed exactly as above , except at a particular step you substitute $x^2=t$
Then the standard result will show up.
I will leave that calculation to yourself.
A: 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!
