Let's solve the limit problem step-by-step.

Problem:

We are tasked with finding the limit of the function as $x \to 1^+$ for the following:

$\lim_{x \to 1^+} \frac{2x^2 - 5x - 3}{x^2 + x - 2}$

Step 1: Direct Substitution

We'll first substitute $x = 1$ into the function directly to check if we get an undefined expression:

$\text{Numerator:} \quad 2(1)^2 - 5(1) - 3 = 2 - 5 - 3 = -6$

$\text{Denominator:} \quad (1)^2 + (1) - 2 = 1 + 1 - 2 = 0$

The denominator equals 0, so we have the indeterminate form $\frac{-6}{0}$. Since the numerator is non-zero, we have an undefined expression at $x = 1$, and this is a limit of the form $\frac{K}{0}$ where $K \neq 0$. In such cases, we need to analyze the behavior of the function near $x = 1$ to determine the limit.

Step 2: Factorize the Function

To simplify the expression, we'll factor both the numerator and the denominator.

Numerator:

The numerator is $2x^2 - 5x - 3$. We need to factor this quadratic expression:

$2x^2 - 5x - 3 = (2x + 1)(x - 3)$

Denominator:

The denominator is $x^2 + x - 2$, which can also be factored:

$x^2 + x - 2 = (x - 1)(x + 2)$

Now, we can rewrite the original expression as:

$\frac{2x^2 - 5x - 3}{x^2 + x - 2} = \frac{(2x + 1)(x - 3)}{(x - 1)(x + 2)}$

Step 3: Analyze the Limit as $x \to 1^+$

At this point, we can analyze the behavior of the function as $x$ approaches 1 from the right.

Substitute $x = 1$ into the factored form:

$\text{Numerator:} \quad (2(1) + 1)(1 - 3) = (2 + 1)(-2) = 3(-2) = -6$

$\text{Denominator:} \quad (1 - 1)(1 + 2) = (0)(3) = 0$

Since the denominator tends to 0 and the numerator remains non-zero, the limit will approach $\pm \infty$. To determine the exact behavior, we need to consider the sign of the denominator as $x \to 1^+$.

As $x \to 1^+$, the denominator:

(x - 1) \to 0^+ \quad \text{(positive side, since x \to 1^+)}

Thus, the denominator approaches 0 from the positive side. The numerator remains negative ($-6$), and so the limit will be:

$\lim_{x \to 1^+} \frac{(2x + 1)(x - 3)}{(x - 1)(x + 2)} = -\infty$

Final Answer:

$\lim_{x \to 1^+} \frac{2x^2 - 5x - 3}{x^2 + x - 2} = -\infty$

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Would you like any further clarification or additional steps? Here are some follow-up questions:

1. How do we factor quadratic expressions efficiently?
2. What does it mean for a limit to approach $\pm \infty$?
3. Can limits of the form $\frac{K}{0}$ have finite values?
4. What happens if both the numerator and denominator tend to zero (indeterminate form $0/0$)?
5. How does the behavior of the function change when we approach from the left, $x \to 1^-$?

Tip: When you encounter indeterminate forms like $\frac{K}{0}$, always analyze the sign of the denominator near the point to determine if the limit tends to $+\infty$ or $-\infty$.