Is $\lim_{x→a} f(x) $ equivalent to $\lim_{h→0} f(a+h)$?

Question

$\lim_{x→a} f(x) = \lim_{h→0} f(a+h)$ does this always hold true? Intuitively these two things seem to describe same concept, and I've seen some algebraic manipulations that implicitly use this identity. Are there any scenarios where this falls apart? Also, if it does hold true, why is that (putting the intuition aside)? Thanks.

Answer 1

Yes, these are equivalent because for every $x$ there is an $h$ (which equals $x-a$), and conversely, for every $h$ there is an $x$ ($=a+h$). It is essentially a renaming of the variables.

The answer would be different for a non bijective change of variable.

$$\lim_{x\to0}\text{sgn}(x)$$ does not exist, while

$$\lim_{t\to0}\text{sgn}(t^2)=1.$$

Answer 2

If $\lim_{x\rightarrow a}f(x)=L$, then by $\epsilon$-argument, somehow it is like $0<|x-a|<\delta$, then $|f(x)-L|<\epsilon$, for $0<|h|<\delta$, then $|(a+h)-a|=|h|$, so $0<|(a+h)-a|<\delta$, set $x=a+h$, then $|f(a+h)-L|=|f(x)-L|<\epsilon$, this shows $\lim_{h\rightarrow 0}f(a+h)=L$. Another way is similar.