Solutions

Let $\lim_{x \to a} f(x)=l$. By definition, we have

$$\forall \epsilon > 0,\ \exists \delta > 0,\ 0<|x-a|<\delta \implies |f(x)-l|<\epsilon.$$

Let $h=x-a$. It follows that $x=a+h$ and

$$\forall \epsilon > 0,\ \exists \delta > 0,\ 0<|h|<\delta \implies |f(a+h)-l|<\epsilon,$$

which is equivalent to

$$\lim_{h \to 0} f(a+h)=l.$$