Formal proof of $\lim_{x\to a}f(x) = \lim_{h\to 0} f(a+h)$ [closed]
Solution 1:
Suppose $\displaystyle \lim_{x \to a}f(x) = c$ and let $\epsilon > 0$, then there is $\delta >0$ such that $$|x-a|<\delta \implies |f(x)-c | < \epsilon$$ now note that $$|h| = |(a+h)-a|$$ and so $$|h|< \delta \implies |f(a+h)-c | < \epsilon$$ which shows that $\displaystyle \lim_{h \to 0}f(a+h) = c$. It follows that $$\displaystyle \lim_{x \to a}f(x) = c= \displaystyle \lim_{h \to 0}f(a+h).$$