Question

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

Answer 1

Let `f(x)=frac{1}{x-3}`, `varepsilon\inRR^+`, `delta=frac{1}{varepsilon}+3`.

`|f(x)-0|<\epsilon` for `x>\delta`, for all `varepsilon`.

Therefore, `lim_{x->oo}f(x)=0`.

Explanation:

Let `f(x)=frac{1}{x-3}`. To say that

`lim_{x->oo}f(x)=0`

means that `f(x)` can be made as close as desired to `0` by making the independent variable `x` close enough to `oo`.

Let the positive number `varepsilon` be how close one wishes to make `f(x)` to `0`. Let `delta` be a real number that denotes how close one will make `x` to `oo`.

The limit exist if, for every `varepsilon>0`, there exist a `delta\inRR` such that

`0-varepsilon<\f(x)<0+varepsilon`

for all `x>delta`.

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

`f(x)<\varepsilon`

The inequality can be simplified to

`x>\frac{1}{varepsilon}+3`

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

`f(x)=frac{1}{x-3}<\frac{1}{delta-3}=varepsilon`