The Definition of the Limit — Question 10

Question 10 - Solution

We are given: $$f(x) = \frac{x^2 - 1}{x - 1}$$ Note that: $$f(x) = \frac{(x - 1)(x + 1)}{x - 1} = x + 1 \quad \text{for } x \neq 1$$

So for $x \neq 1$, $f(x) = x + 1$, and: $$\lim_{x \to 1} f(x) = \lim_{x \to 1} (x + 1) = 2$$

We now prove this using the $\varepsilon$-$\delta$ definition.

We want to show: $$\forall \varepsilon > 0, \exists \delta > 0 \text{ such that } 0 < |x - 1| < \delta \Rightarrow \left| f(x) - 2 \right| < \varepsilon$$

Since $f(x) = x + 1$ for $x \neq 1$, we compute: $$|f(x) - 2| = |x + 1 - 2| = |x - 1|$$

So: $$|f(x) - 2| < \varepsilon \quad \text{whenever} \quad |x - 1| < \varepsilon$$

Step: Choose $\delta = \varepsilon$

Then if $0 < |x - 1| < \delta$, we get: $$|f(x) - 2| = |x - 1| < \varepsilon$$

Conclusion: By the $\varepsilon$-$\delta$ definition, we have: $$\boxed{\lim_{x \to 1} \frac{x^2 - 1}{x - 1} = 2}$$