Proving that $x/(x+1)$ is continuous at $x=2$ by the definition

Question

I came upon this question,

the function $f(x)=\frac{x}{x+1}$ is continuous at $x=2$ by the definition.

Do I have to show that limit at $x=2$ exits first? I am confused with the steps, can someone please help?

Answer 1

Write $f(x)=1-\frac{1}{x+1}$ so that $$ f(x)-f(2)=\frac{1}{3}-\frac{1}{x+1}=\frac{x-2}{3(x+1)}. $$ Note that for $x$ sufficiently close to $2$, we have $x+1>1$. So for all such $x$, $|f(x)-f(2)|<\frac{|x-2|}{3}$.

The observations above should be enough for you to show $f(x)\to f(2)$ as $x\to 2$. But if you get stuck:

$$\forall e>0,\exists\delta=\min\{2,3e\}>0:|x-2|<\delta\implies|f(x)-f(2)|<e.$$