How can I prove by epsilon-delta definition that the limit of the following function is $\frac{4}{5}$.

Question

How can I prove by epsilon-delta definition the following : $$\begin{equation*} \lim_{x \rightarrow 2} \frac{x^3 - 4}{x^2 +1}= \frac{4}{5} \end{equation*}$$

Shall I use long division? any help will be appreciated.

Answer 1

I would use long division to begin with, yes. Note that, $$\frac{x^3 - 4}{x^2 + 1} = x - \frac{x + 4}{x^2 + 1}.$$ I also often find it's often helpful to transpose the problem so that $x \rightarrow 0$. We take instead $u = x - 2 \iff x = u + 2$. Then the problem becomes: $$\lim_{u \rightarrow 0} u + 2 - \frac{u + 6}{u^2 + 4u + 5} = \frac{4}{5}.$$

Fix $\varepsilon > 0$. We have \begin{align*} & \left| u + 2 - \frac{u + 6}{u^2 + 4u + 5} - \frac45 \right| < \varepsilon \\ \iff& \left| u - \frac{u + 6}{u^2 + 4u + 5} + \frac65 \right| < \varepsilon \\ \impliedby& |u| + \left|\frac{u + 6}{u^2 + 4u + 5} - \frac65 \right| < \varepsilon \\ \iff& |u| + \left|\frac{5(u + 6)}{5(u^2 + 4u + 5)} - \frac{6(u^2 + 4u + 5)}{5(u^2 + 4u + 5)} \right| < \varepsilon \\ \iff& |u| + |u| \cdot \left|\frac{6u + 19}{5(u^2 + 4u + 5)} \right| < \varepsilon. \end{align*} Note that $u^2 + 4u + 5 = x^2 + 1$ has a minimum value of $1$ globally. We can also bound $6u + 19$ locally by forcing $|u| < 1$ (or indeed any other positive number). Under this assumption, we have $|6u + 19| \le 6|u| + 19 < 25$, and so, $$\impliedby |u| + 25|u| < \varepsilon \text{ and }|u| < 1.$$ Therefore, we may take $\delta = \min\left\lbrace 1, \frac{\varepsilon}{26} \right\rbrace$. The same $\delta$ will work for the problem in $x$ as well.

Answer 2

Just Plug It In

Since the denominator doesn't vanish as $x\to2$, and $(u,v)\mapsto\frac uv$ is continuous as long as $v$ doesn't vanish, we can just evaluate the ratio at $x=2$: $$ \begin{align} \lim_{x\to2}\frac{x^3-4}{x^2+1} &=\left.\frac{x^3-4}{x^2+1}\right|_{x=2}\\ &=\frac{8-4}{4+1}\\[6pt] &=\frac45 \end{align} $$

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Use $\boldsymbol{\epsilon\text{-}\,\delta}$

Since we are looking at the limit as $x\to2$, we can assume $1\le x\le3$. Then we have $\left|\,5x^2+6x+12\,\right|\le75$ and $5x^2+5\ge10$, so if $|x-2|\le\epsilon$ $$ \begin{align} \delta &=\left|\frac{x^3-4}{x^2+1}-\frac45\right|\\ &=\left|\frac{5x^3-4x^2-24}{5x^2+5}\right|\\ &=\left|\frac{5x^2+6x+12}{5x^2+5}\right|\,|x-2|\\[3pt] &\le\frac{15}2\,\epsilon \end{align} $$