To solve the limit \( \lim_{x \to 1} \frac{2x^2}{x^2 - 1} \), we can approach it step-by-step:

1. **Direct Substitution**: First, substitute \( x = 1 \):
\[
\frac{2(1)^2}{(1)^2 - 1} = \frac{2}{1 - 1} = \frac{2}{0}
\]
This is undefined, which suggests further analysis is needed.

2. **Factor the Denominator**: The expression in the denominator can be factored:
\[
x^2 - 1 = (x - 1)(x + 1)
\]
Thus, the limit can be rewritten:
\[
\lim_{x \to 1} \frac{2x^2}{(x - 1)(x + 1)}
\]

3. **Simplifying the Expression**: Since direct substitution leads to a form that is undefined, we check for any cancellations. There are no factors of \( x - 1 \) in the numerator to cancel, so we will need to analyze the behavior of the limit as \( x \) approaches 1.

4. **Using L'Hôpital's Rule**: Since we have an indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule:
Differentiate the numerator and denominator:
- Numerator: \( \frac{d}{dx}(2x^2) = 4x \)
- Denominator: \( \frac{d}{dx}((x - 1)(x + 1)) = \frac{d}{dx}(x^2 - 1) = 2x \)

5. **Applying L'Hôpital's Rule**:
\[
\lim_{x \to 1} \frac{4x}{2x} = \lim_{x \to 1} \frac{4}{2} = 2
\]

Thus, the limit is:
\[
\boxed{2}
\]