# Evaluate $\lim_{x \to +\infty}\frac{2x}{(x-3)}$

Evaluate the limit as $\lim_{x \to +\infty}$ and prove the result: $f(x) = \frac{2x}{(x-3)}$

I know that the limit is $2$ but having a hard time to find the right $\epsilon$. Please help

To show a limit, you should not find the $\epsilon$. Rather you will be provide with any $\epsilon$ and your goal, in this case, is to find a value $M(\epsilon)$ such that for $x>M(\epsilon)$, we have $$\left\vert \dfrac{2x}{x-3} - 2 \right\vert < \epsilon$$ To find a candidate $M(\epsilon)$, let us consider $\dfrac{2x}{x-3} - 2$. We have $$\dfrac{2x}{x-3} - 2 = \dfrac6{x-3}$$ To make this less than $\epsilon$, one candidate $M(\epsilon)$ is $3+\dfrac6{\epsilon}$. Now check if this candidate does indeed satisfy our expectation, i.e., our definition of the limit.