To prove the precise definition of the limit statement:

$\lim_{{x \to 6}} \frac{1}{x} = \frac{1}{6}$

we need to use the ε-δ definition of a limit. According to this definition:

For every $\epsilon > 0$, there must exist a $\delta > 0$ such that for all $x$ satisfying $0 < |x - 6| < \delta$, it follows that $\left| \frac{1}{x} - \frac{1}{6} \right| < \epsilon$.

Step-by-Step Proof:

1. Start with the inequality $\left| \frac{1}{x} - \frac{1}{6} \right| < \epsilon$.

We want to manipulate this inequality to express $|x - 6|$ in terms of $\epsilon$.

$\left| \frac{1}{x} - \frac{1}{6} \right| = \left| \frac{6 - x}{6x} \right|$

This simplifies to:

$\left| \frac{6 - x}{6x} \right| < \epsilon$

2. Isolate $|6 - x|$:

Multiply both sides by $6x$:

$|6 - x| < 6x\epsilon$

3. Find bounds on $x$:

Next, we place some bounds on $x$ so that $6x$ stays controlled. Since we're considering $x$ near 6, let's assume $|x - 6| < 1$, which implies:

$5 < x < 7$

For $x$ in this interval, the value of $6x$ satisfies:

$30 < 6x < 42$

Thus, we can say that $6x$ is bounded below by 30. This gives us:

$|6 - x| < 6x\epsilon \quad \Rightarrow \quad |6 - x| < 30\epsilon$

4. Choose $\delta$:

To ensure that $\left| \frac{1}{x} - \frac{1}{6} \right| < \epsilon$, we must choose $\delta$ so that $|x - 6| < \delta$ implies $|x - 6| < 30\epsilon$. Thus, we set:

$\delta = 30\epsilon$

Conclusion:

For any $\epsilon > 0$, if we choose $\delta = 30\epsilon$, then for all $x$ such that $0 < |x - 6| < \delta$, it follows that:

$\left| \frac{1}{x} - \frac{1}{6} \right| < \epsilon$

This completes the proof using the ε-δ definition of a limit.

Would you like further details on any step, or perhaps some additional examples? Here are some related questions for further exploration:

1. How does the ε-δ definition of a limit work for functions like $\frac{1}{x}$?
2. What is the geometric interpretation of the ε-δ definition of a limit?
3. How do you handle limits involving infinity with the ε-δ definition?
4. How can the δ value change based on different functions?
5. What happens if you pick different bounds for $x$ in the proof?

Tip: When working with limits, simplifying the expression within the limit is often key to identifying how $\delta$ depends on $\epsilon$.