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

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.
