I'm trying to use the $\varepsilon$–$\delta$ definition of a limit to prove that
$$
\lim_{x \to 1} \frac{1}{x+2} = 1/3.
$$
But I'm getting stuck on finding the correct $\delta$. Here is my try:
\begin{align*}
\lvert f(x) – L \rvert < \varepsilon \\
\lvert \frac{1}{x+2} – \frac{1}{3} \rvert < \varepsilon \\
\lvert \frac{x -1}{2 + x} \rvert < 3\varepsilon.
\end{align*}
And then I'm not really sure what to do. How do you proceed from here?
Best Answer
So far you have deduced a chain of equivalences, and I will add one more to the chain: \begin{align*} & \lvert f(x) - L \rvert < \varepsilon \\ \iff & \lvert \frac{1}{x+2} - \frac{1}{3} \rvert < \varepsilon \\ \iff & \lvert \frac{x -1}{2 + x} \rvert < 3\varepsilon \\ \iff & \lvert x-1 \rvert < 3 \, \epsilon \, \lvert 2+x \rvert \end{align*} Reading this chain of equivalences backwards, you have proved so far that $$\lvert x-1 \rvert < 3 \, \epsilon \, \lvert 2+x \rvert \implies \lvert f(x) - L \rvert < \varepsilon $$
Your goal now is to find a value of $\delta>0$ such that $$ \lvert x-1 \rvert < \delta \implies \lvert x-1 \rvert < 3 \, \epsilon \, \lvert 2+x \rvert $$ for then you will be done.
The first step towards finding $\delta$ is to work with the factor $\lvert 2+x \rvert$ separately. Your goal in this step is to find any lower bound to that factor. Intuitively, if $x$ is close to $1$ then it must be far from $-2$. To make this intuition rigorous, note that $$\lvert x-1 \rvert < 1 \implies -1 < x-1 < +1 \implies 2 < x+2 < 4 \implies 2 < \lvert x+2 \rvert $$ Summarizing, we have found $\delta_1 = 1$ for which the following implications hold: $$\lvert x-1 \rvert < \delta_1 \implies \lvert x-1 \rvert < 1 \implies 2 < \lvert 2+x \rvert \implies 6 \epsilon < 3 \epsilon \lvert 2+x \rvert $$ Now we set $\delta_2 = 6 \epsilon$.
Assuming already that $\lvert x-1 \rvert < \delta_1 = 1$, if in addition we assume that $\lvert x-1 \rvert < \delta_2 = 6 \epsilon$ then it follows that $$\lvert x-1 \rvert < \delta_2 = 6 \epsilon < 3 \epsilon \lvert 2+x \rvert $$ And now we have found $\delta$, namely: $$\delta = \min\{\delta_1,\delta_2\} $$ Putting this altogether, assuming that $\lvert x-1 \rvert < \delta$ it follows that $\lvert x-1 \rvert < \delta_1 = 1$ and $\lvert x-1 \rvert < \delta_2 = 6 \epsilon$ and therefore $$\lvert x-1| < 3 \epsilon \lvert x+2 \rvert $$ This, as we have already shown, implies that $\lvert f(x)-L \rvert < \epsilon$.