# How do you use the formal definition of a limit to find 1/(x - 3) = 0 as x approaches infinity?

Nov 8, 2015

Let $f \left(x\right) = \frac{1}{x - 3}$, $\varepsilon \setminus \in {\mathbb{R}}^{+}$, $\delta = \frac{1}{\varepsilon} + 3$.

$| f \left(x\right) - 0 | < \setminus \epsilon$ for $x > \setminus \delta$, for all $\varepsilon$.

Therefore, ${\lim}_{x \to \infty} f \left(x\right) = 0$.

#### Explanation:

Let $f \left(x\right) = \frac{1}{x - 3}$. To say that

${\lim}_{x \to \infty} f \left(x\right) = 0$

means that $f \left(x\right)$ can be made as close as desired to $0$ by making the independent variable $x$ close enough to $\infty$.

Let the positive number $\varepsilon$ be how close one wishes to make $f \left(x\right)$ to $0$. Let $\delta$ be a real number that denotes how close one will make $x$ to $\infty$.

The limit exist if, for every $\varepsilon > 0$, there exist a $\delta \setminus \in \mathbb{R}$ such that

$0 - \varepsilon < \setminus f \left(x\right) < 0 + \varepsilon$

for all $x > \delta$.

We already know that $f \left(x\right) > 0 > 0 - \varepsilon$ for all $x > 3$. All that is left is the upper bound.

$f \left(x\right) < \setminus \varepsilon$

The inequality can be simplified to

$x > \setminus \frac{1}{\varepsilon} + 3$

Let $\delta = \frac{1}{\varepsilon} + 3$. We can see that for all $x > \delta \left(> 3\right)$,

$f \left(x\right) = \frac{1}{x - 3} < \setminus \frac{1}{\delta - 3} = \varepsilon$