Limit of sum/difference condition which are not being followed here This was given as an example in a book : If $$\lim _{x \rightarrow a}[f(x)+g(x)]=2$$ and
$$\lim _{x \rightarrow a}[f(x)-g(x)]=1,$$ then find the value of $$\lim _{x \rightarrow a} f(x) g(x)$$
Solution :  $$\lim _{x \rightarrow a}[f(x)+g(x)]=2$$
or $$\lim _{x \rightarrow a} f(x)+\lim _{x \rightarrow a} g(x)=2$$ $\tag{1}$
$$\lim _{x \rightarrow a}[f(x)-g(x)]=1$$
or $$\lim _{x \rightarrow a} f(x)-\lim _{x \rightarrow a} g(x)=1$$ $\tag{2}$
Adding (1) and (2),
$$2 \lim _{x \rightarrow a} f(x)=3$$ or $$\lim _{x \rightarrow a} f(x)=\frac{3}{2}$$
Subtracting (2) from (1),
$$2 \lim _{x \rightarrow a} g(x) \neq 1$$ or $$\lim _{x \rightarrow a} g(x)=\frac{1}{2}$$
or $$\lim _{x \rightarrow a} f(x) g(x)=\lim _{x \rightarrow a} f(x) \lim _{x \rightarrow a} g(x)=\frac{3}{2} \times \frac{1}{2}=\frac{3}{4}.$$

My query is that isnt the lim of sum = individual limits sum only when we  already know that limit of individual sum  exists ? Here we actually dont know if it actually exists or not so shouldnt this method is totally wrong ?


You are right that the way it is written is slightly problematic. That $\lim_{x\to a} f(x)+g(x)$ exists doesn't mean that $\lim_{x\to a} f(x)$ and $\lim_{x\to a} g(x)$ exist. However, there is no actual reason to split the limits and try writing
\begin{align*}
\lim _{x \rightarrow a} f(x)+\lim _{x \rightarrow a} g(x)=2\\
\lim _{x \rightarrow a} f(x)-\lim _{x \rightarrow a} g(x)=1
\end{align*}
as they have. Instead, work directly with
\begin{align*}
\lim_{x\to a} f(x)+g(x)=2\\
\lim_{x\to a} f(x)-g(x)=1
\end{align*}
and add these two together (as the limit of a sum is the sum of the limits), and get
$$\lim_{x\to a} f(x)+g(x)+f(x)-g(x)=3$$
i.e.
$$\lim_{x\to a} 2f(x)=3$$
and indeed $\lim_{x\to a} f(x)$ exists, while no unjustified splitting of limits has occurred.
