$$\lim_{x \to 0}\left(\dfrac 1 {\sin x}-\dfrac1 x\right) $$
I solved this problem using the fact that near $x=0$ , $\sin x \cong x$),$(\sin x=x+O(x^2)$ therefore subtracting them results to $0$ which I did verify using other techniques, However, the same reasoning leads to $0$ in solving
$$\lim_{x \to 0}\left(\dfrac 1 {\log (x+1)}-\dfrac1 x\right) $$
while the correct result is $1/2$ although $\log(x+1) \cong x$ when $x$ is near zero$(\log (x+1)=x+O(x^2)$.What is getting wrong here?
On
We will be using the following three Taylor expansions: $\sin{x}=x-\frac{x^3}{6}+O(x^5)$, $\log(1+x)=x-\frac{x^2}{2}+O(x^3)$ and $\frac{1}{1-x}=1+x+O(x^2)$
$$\begin{align}\frac{1}{\sin{x}}&=\frac{1}{x-\frac{x^3}{6}+O(x^5)}\\&=\frac{1}{x(1-x^2/6+O(x^4))}\\&=\frac{1}{x}(1+\frac{x^2}{6}+O(x^4))\\&=\frac{1}{x}+\frac{x}{6}+O(x^3)\end{align}$$
And the limit in this case is zero. While
$$\begin{align}\frac{1}{\log(1+x)}&=\frac{1}{x-\frac{x^2}{2}+O(x^3)}\\&=\frac{1}{x(1-x/2+O(x^2))}\\&=\frac{1}{x}(1+\frac{x}{2}+O(x^2))\\&=\frac{1}{x}+\frac{1}{2}+O(x)\end{align}$$
And the limit is $\frac{1}{2}$
On
Your error comes from the fact that equivalence of functions is not compatible with addition and substraction.
Counter-example:
We have $x+x^2\sim_0 x$, $\,-x\sim_0 -x+x^3$, but $$(x+x^2)-x=x^2 \not\sim_0 x+(-x+x^3)=x^3. $$
While equivalence is a powerful tool for computing limits as it frees from irrelevant computational details, it must be used cautiously:
Main rules for computing with equivalence:
Let $f,g$ functions defined in a neighbourhood of $a \in \overline{\mathbf R}$. If $f\sim_a g $, $f_1\sim_a g_1$, then:
On
This is one of the most common problems associated with the use of symbol $\sim$ (or here $\cong$ as used by OP).
The error comes because of a lack of proper understanding of the definition of symbol $\sim$. The proper definition of this symbol is as follows:
If $\lim\limits_{x \to a}\dfrac{f(x)}{g(x)} = 1$ then we write $f(x) \sim g(x)$ as $x \to a$.
From the above definition it should be clear enough (although many students almost always fail to get it) that this definition is in a multiplicative context. Thus if $f(x) \sim g(x)$ as $x \to a$ then we can replace $f(x)$ by $g(x)$ while calculating the limit of an expression containing $f(x)$ provided $f(x)$ is in a multiplicative context in that expression. Here "multiplicative context" means that the expression containing $f(x)$ must be of type $f(x)h(x)$ (or like $h(x)/f(x)$). And then we can as well calculate the limit of $g(x)h(x)$ and get answer.
This is because $$\lim_{x \to a}f(x)h(x) = \lim_{x \to a}\frac{f(x)}{g(x)}\cdot g(x)h(x) = \lim_{x \to a}1\cdot g(x)h(x)$$
It is in this manner that we replace $f(x)$ by $g(x)$ when $f(x) \sim g(x)$. When calculating limits one must know exactly which rules of limits are used in each step (even when they are being used implicitly).
In the current context $\log(1 + x) \sim x$ as $x \to 0$, but expression $$\left(\frac{1}{\log(1 + x)} - \frac{1}{x}\right)$$ is not of the form $\log(1 + x)\cdot \text{(some function)}$. Well you may write the expression as $$\frac{1}{\log(1 + x)}\left(1 - \frac{\log(1 + x)}{x}\right)$$ and the first factor definitely be replaced by $1/x$ to get $$\frac{1}{x}\left(1 - \frac{\log(1 + x)}{x}\right)$$
Note: Many people try to complicate the things by adding the talk of "order". One should not attach any more meaning to the statement $f(x) \sim g(x)$ than is provided by its definition. Trying to attach more meaning to this statement like "$f(x)$ is almost equal to $g(x)$ to first order of smallness" is only going to add confusion and not make it more intuitive.
The first try fails because the limit of both terms is $\infty$. You need the stronger result $$ \sin x = x + o(x^2) $$
and then you get $$ \frac 1{\sin x} - \frac 1x = \frac{x - \sin x}{x\sin x} = \frac{o(x^2)}{x\sin x} = o(1)\to 0 $$
More generally if $$ f(x) = x + a x^2 + o(x^2) $$(important case: $f$ is twice differentiable around 0 with $a = f''(0)/2$) then the limit is $-a$.