Changing Sides of Limits Property Is the following property about limits correct?$$\lim_{x+b\to a}\,f(x)=\lim_{x\to a-b}\,f(x)$$
and does it also mean $$\lim_{x\to a}\,f(x)=\lim_{nx\to na}\,f(x)=\lim_{x^2\to a^2}\,f(x)=\lim_{g(x)\to g(a)}\,f(x)$$ etc.?
I encountered it when trying to show $f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$.
Intuitively it seems true and similar to manipulating $x=a$ but I haven't found anything when searching online about it.
 A: Actually the notation of limit is always of the form $\lim_{x \to a}f(x)$ or in somewhat crude terms $$\lim_{\text{variable }\to \text{ some value}}\text{ function of the variable}$$ You should not write something like $$\lim_{g(x) \to b}f(x)$$ because although the meaning is intuitively clear enough it is still an abuse of the limit notation.

What you are trying to express in your question is better handled via rule of substitution.
Rule of substitution: Let $f$ be defined in a certain deleted neighborhood of $a$ and let $\lim_{x \to a}f(x) = L$. Let $g$ be another function defined in a deleted neighborhood of $b$ such that $g(t) \neq a$ for all values of $t$ in this specific neighborhood of $b$ and further let $\lim_{t \to b}g(t) = a$ then $$\lim_{x \to a}f(x) = L = \lim_{t \to b}f(g(t))$$
The rule also holds if one or both of $a, L$ are infinite. It is important to note that the converse of the above theorem does not hold in general so that if $\lim_{t \to b}f(g(t)) = L$ and $\lim_{t \to b}g(t) = a$ exist and $g(t) \neq a$ in a certain deleted neighborhood of $b$ then it does not necessarily follows that $\lim_{x \to a}f(x)$ exists.
Now consider the limit $$\lim_{x \to a}\frac{f(x) - f(a)}{x - a}$$ and consider another function $g$ given by $g(h) = a + h$. Clearly $\lim_{h \to 0}g(h) = a$ and further $g(h) \neq a$ for all values of $h$ in any deleted neighborhood of $0$. Hence the rule of substitution applies and we have $$\lim_{x \to a}\frac{f(x) - f(a)}{x - a} = \lim_{h \to 0}\frac{f(g(h)) - f(a)}{g(h) - a} = \lim_{h \to 0}\frac{f(a + h) - f(a)}{h}$$

However the equality $$\lim_{x \to a}f(x) = L = \lim_{h \to 0}f(a + h)\tag{1}$$ is more easily proved by a direct application of the definition of limit rather than using the rule of substitution mentioned above. When you try to express both the limits in $(1)$ in terms of $\epsilon, \delta$ and compare both the definitions you will find that they are exactly same word for word except that the implication $$0 < |x - a| < \delta \Rightarrow |f(x) - L| < \epsilon$$ is replaced by $$0 < |h| < \delta \Rightarrow |f(a + h) - L| < \epsilon$$ and these two implications are same if you actually look at the values taken by the argument of $f$. Thus the equality $(1)$ is a direct and immediate consequence of the definition of limits and one should not be required to prove it unless one is specifically doing some exercise on definition of limits.
A: the first property($\lim_{x\to 0^+}\,f(x)=\lim_{x\to 0^-}\,f(x)$ )is not always true but depends on the continuity of the function at the point where the limit is going to be applied.
for example: the limit will not be true for a signum(x) function at x=0 as the function is discontinuous and the left hand limit($\lim_{x\to0^+}$ will be -1 while the right hand limit($\lim_{x\to0^-}$) is +1 but actual value at x=0 is 0
A: 1).Suppose $\lim_{x+b\to a}\,f(x)=L$ You want to show that $\lim_{x\to a-b}\,f(x)=L$ also. So, let $\epsilon >0$ be given. 
Then there is a $\delta >0$ such that $\vert f(x)-L\vert <\epsilon$ whenever $\vert a-(x+b)\vert<\delta,\ $which is the same thing as $\vert (a-b)-x\vert $ so using the same $\delta$ we see that the second limit is also $L$ and we are done. 
2). If $g$ is not continuous, this may fail (why?). On the other hand, suppose that $\lim_{g(x)\to g(a)}\,f(x)=L$ and $g$ is continuous. Then there is a $\delta >0$ such that 
$\tag1\vert g(x)-g(a)\vert <\delta\Rightarrow \vert f(x)-L\vert<\epsilon $ 
and for this $\delta$ there is an $\eta >0$ such that 
$\tag2 \vert x-a\vert <\eta \Rightarrow \vert g(x)-g(a)\vert <\delta$
Combining $(1)$ and $(2)$ gives us what we need. 
