1. We are asked to prove that \(\lim_{x \to 1} f(x) = 1\) where \(f(x) = \begin{cases} x^2, & x \neq 1 \\ 2, & x = 1 \end{cases}\). 2. We want to show for every \(\varepsilon > 0\), there exists \(\delta > 0\) such that if \(0 < |x - 1| < \delta\), then \(|f(x) - 1| < \varepsilon\). 3. Since \(f(x) = x^2\) when \(x \neq 1\), we consider \(|f(x) - 1| = |x^2 - 1| = |x-1||x+1|\). 4. To control \(|x+1|\), note if \(|x - 1| < 1\), then \(x \in (0, 2)\) so \(|x + 1| \leq 3\). 5. Therefore, \(|x^2 - 1| = |x-1||x+1| < 3|x-1|\). 6. To have \(|x^2 - 1| < \varepsilon\), it suffices that \(3|x-1| < \varepsilon \implies |x-1| < \frac{\varepsilon}{3}\). 7. Set \(\delta = \min(1, \frac{\varepsilon}{3})\) so that both conditions are satisfied. 8. Hence, for \(0 < |x-1| < \delta\), we get \(|f(x) - 1| < \varepsilon\), proving the limit. 9. The value of \(f(1) = 2\) does not affect the limit since limits depend on values arbitrarily close but not equal to \(x=1\). 10. Therefore, \(\boxed{\lim_{x \to 1} f(x) = 1}\).