To prove that $\lim_{x \to 1} \frac{1}{x} = 1$ using the epsilon-delta definition of a limit, we start with the formal definition:

For every $\epsilon > 0$, there exists a $\delta > 0$ such that if $0 < |x - 1| < \delta$, then $|\frac{1}{x} - 1| < \epsilon$.

Step 1: Express the Limit Condition

We want to show:

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

This simplifies to:

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

Step 2: Manipulate the Inequality

Rearranging gives:

$\frac{|1 - x|}{|x|} < \epsilon$

This implies:

$|1 - x| < \epsilon |x|$

Step 3: Choose $\delta$

We want to find a suitable $\delta$ such that if $0 < |x - 1| < \delta$, then $x$ is close to 1. We can bound $|x|$:

If $x$ is close to 1, specifically, if $|x - 1| < \frac{1}{2}$, then:

$\frac{1}{2} < x < \frac{3}{2}$

Thus, $|x|$ is bounded away from zero. In this case, we have:

$\frac{1}{2} < |x| < \frac{3}{2}$

Step 4: Finalizing $\epsilon$

Now, since $|x| \geq \frac{1}{2}$, we can replace $|x|$ in our inequality:

$|1 - x| < \epsilon |x| \implies |1 - x| < 2\epsilon$

Step 5: Set $\delta$

Let $\delta = \min\left(\frac{1}{2}, 2\epsilon\right)$. Therefore, if $0 < |x - 1| < \delta$, then:

$|1 - x| < 2\epsilon \implies \left| \frac{1}{x} - 1 \right| < \epsilon$

Conclusion

We have shown that for every $\epsilon > 0$, we can find a $\delta > 0$ such that:

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

Questions to Consider

1. How does the choice of $\delta$ affect the proof?
2. Can this method be used for other limits?
3. What would change if we proved a limit as $x \to 0$?
4. How does the concept of continuity relate to limits?
5. What are common pitfalls in using epsilon-delta definitions?

Tip: Always visualize the limit on a graph to understand the behavior of the function near the limit point. Would you like more details on any part of this proof?