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The given expression is:

\[
\lim_{{x \to 1}} \left( \frac{x(x + 1)}{x^2 - 1} - \frac{2}{x^2 - 1} \right)
\]

Let's solve it step by step.

### Step 1: Simplify the Denominator
Notice that the denominator \(x^2 - 1\) can be factored:

\[
x^2 - 1 = (x - 1)(x + 1)
\]

Thus, the expression becomes:

\[
\lim_{{x \to 1}} \left( \frac{x(x + 1)}{(x - 1)(x + 1)} - \frac{2}{(x - 1)(x + 1)} \right)
\]

### Step 2: Combine the Terms
Since both terms have the same denominator, we can combine them into a single fraction:

\[
\lim_{{x \to 1}} \frac{x(x + 1) - 2}{(x - 1)(x + 1)}
\]

### Step 3: Simplify the Numerator
Now, simplify the numerator \(x(x + 1) - 2\):

\[
x(x + 1) - 2 = x^2 + x - 2
\]

Therefore, the expression becomes:

\[
\lim_{{x \to 1}} \frac{x^2 + x - 2}{(x - 1)(x + 1)}
\]

### Step 4: Factor the Numerator
Notice that \(x^2 + x - 2\) can be factored as:

\[
x^2 + x - 2 = (x - 1)(x + 2)
\]

So the expression becomes:

\[
\lim_{{x \to 1}} \frac{(x - 1)(x + 2)}{(x - 1)(x + 1)}
\]

### Step 5: Cancel the Common Factor
We can cancel the common factor of \((x - 1)\) from the numerator and the denominator:

\[
\lim_{{x \to 1}} \frac{x + 2}{x + 1}
\]

### Step 6: Evaluate the Limit
Now, substitute \(x = 1\) into the simplified expression:

\[
\frac{1 + 2}{1 + 1} = \frac{3}{2}
\]

### Final Answer:
The value of the limit is:

\[
\boxed{\frac{3}{2}}
\]

Would you like more details, or do you have any further questions?

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Here are five related questions you can explore:
1. What happens if we change the limit point from \(x \to 1\) to another value?
2. How would the solution change if the denominator was not factorable?
3. What is the significance of factoring when solving limits involving polynomials?
4. How do you handle limits that result in indeterminate forms like \(\frac{0}{0}\)?
5. Can L'Hopital's Rule be used to solve this limit? If so, how?

**Tip**: Always check if a limit expression simplifies before directly substituting the value of \(x\).

How to solve the limit equation provided in the image.

Learn how to solve the limit problem involving a rational expression with factoring techniques. This problem covers key calculus concepts like limits and polynomial factoring. Suitable for high school students in Grades 11-12.

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