Evaluate the following limit.
 
Enter the exact answer. If the limit does not exist, enter DNE.
 
limx→1+x2x2−1=


To evaluate the limit

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

we first analyze the expression by direct substitution of \(x = 1\):

\[
\text{Numerator: } 1^2 = 1
\]

\[
\text{Denominator: } 1^2 - 1 = 1 - 1 = 0
\]

This gives us a situation where the numerator is 1 and the denominator is 0. This implies the limit may approach infinity or negative infinity, depending on the direction from which \(x\) approaches 1.

### Step 1: Consider the direction \(x \to 1^+\)
Since \(x \to 1^+\) means \(x\) is slightly greater than 1, let's substitute \(x = 1 + \epsilon\) where \(\epsilon\) is a small positive number.

Then:

\[
\text{Denominator: } (1 + \epsilon)^2 - 1 = 1 + 2\epsilon + \epsilon^2 - 1 = 2\epsilon + \epsilon^2
\]

For very small \(\epsilon\), \(2\epsilon + \epsilon^2\) is positive and close to zero.

So, as \(\epsilon\) approaches 0 from the positive side:

\[
\frac{x^2}{x^2 - 1} = \frac{(1 + \epsilon)^2}{(1 + \epsilon)^2 - 1} \to \frac{1}{0^+} \to +\infty
\]

### Conclusion
As \(x\) approaches 1 from the right (\(x \to 1^+\)), the denominator approaches 0 from the positive side, and since the numerator is positive, the limit tends to \(+\infty\).

Thus, the limit is:

\[
\boxed{\infty}
\]

Would you like to dive deeper into this problem or explore other related limits? Here are some related questions:

1. How do you evaluate limits that approach from the left (\(x \to 1^-\))?
2. What is the difference between \(0^+\) and \(0^-\) in limits?
3. How do limits behave when the denominator approaches zero?
4. What is the significance of the sign of the denominator in one-sided limits?
5. When do limits not exist (DNE) in terms of one-sided limits?

**Tip:** Always consider the direction of approach (from the left or right) when dealing with limits involving \(0\) in the denominator.

Learn how to evaluate the limit lim x→1+ x^2 / (x^2 - 1) step-by-step. Understand the behavior of limits as x approaches 1 from the right (x → 1+). This problem explores one-sided limits and the concept of infinity in calculus.

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