Solutions

Suppose there is a $\delta > 0$ such that $f(x)=g(x)$ when $0<|x-a|<\delta$. Without loss of generality, let $\lim_{x \to a} f(x)=l$. Choose any $\epsilon > 0$. By definition, there exists $\delta' > 0$ such that

$$0<|x-a|<\delta' \implies |f(x)-l|<\epsilon.$$

Let $\delta''=\min(\delta',\delta)$. We thus have

$$0<|x-a|<\delta'' \implies |f(x)-l|<\epsilon \quad\text{and}\quad |g(x)-l|<\epsilon.$$

Hence,

$$\lim_{x \to a} f(x) = \lim_{x \to a} g(x) = l.$$