Answer
See explanations.
Work Step by Step
Step 1. For $\lim_{x\to c}f(x)=L$ to be true, for every number $\epsilon\gt0$, there must exist a corresponding number $\delta\gt0$ such that for all $x$, $0\lt |x-c| \lt \delta$, we get $|f(x)-L|\lt\epsilon$.
Step 2. The last inequality gives $-\epsilon\lt f(x)-L \lt \epsilon$ or $L-\epsilon\lt f(x) \lt L+\epsilon$
Adn the requirement for $\delta$ gives $-\delta \lt x-c \lt \delta$ or $c-\delta \lt x \lt c+\delta$
Step 3. Let $x=h+c$, we have $h=x-c$ and when $x\to c, h\to 0$
Step 4. The limit $\lim_{h\to0}f(h+c)=L$ is true if and only if for every number $\epsilon\gt0$, there exists a corresponding number $\delta\gt0$ such that for all $h$, $0\lt |h| \lt \delta$, we get $|f(h+c)-L|\lt\epsilon$.
Step 5. The inequalities in step 4 lead to $-\epsilon\lt f(h+c)-L \lt \epsilon$ or $L-\epsilon\lt f(h+c) \lt L+\epsilon$
And the requirement for $\delta$ gives $-\delta \lt h \lt \delta$
Step 6. Use the relation in step 3, the above conditions become $L-\epsilon\lt f(x) \lt L+\epsilon$
and $\delta$ gives $-\delta \lt x-c \lt \delta$ or $c-\delta \lt x \lt c+\delta$
Step 7. As the conditions in step 6 is the same of the requirements in step 2, we conclude that the limit $\lim_{x\to c}f(x)=L$ is true if and only if the limit $\lim_{h\to0}f(h+c)=L$ is true.
