Evaluate `lim_(xto1) ((2x-3)(sqrt(x)-1))/(2x^(2)+x-3).`

Explanation

To evaluate the limit

\[ \lim_{x \to 1} \frac{(2x-3)(\sqrt{x}-1)}{2x^2+x-3}, \]

we will follow these steps:

### Step 1: Substitute \( x = 1 \)

First, let's substitute \( x = 1 \) into the expression to check if we get an indeterminate form:

\[ \text{Numerator: } (2(1)-3)(\sqrt{1}-1) = (2-3)(1-1) = (-1)(0) = 0. \]
\[ \text{Denominator: } 2(1)^2 + 1 - 3 = 2 + 1 - 3 = 0. \]

Since both the numerator and denominator evaluate to 0, we have the indeterminate form \( \frac{0}{0} \).

### Step 2: Factor the Denominator

Next, we need to factor the denominator \( 2x^2 + x - 3 \). We can use the quadratic formula to find the roots:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-3)}}{2 \cdot 2} = \frac{-1 \pm \sqrt{1 + 24}}{4} = \frac{-1 \pm 5}{4}. \]

This gives us the roots:

\[ x = 1 \quad \text{and} \quad x = -\frac{3}{2}. \]

Thus, we can factor the denominator as:

\[ 2x^2 + x - 3 = 2(x - 1)(x + \frac{3}{2}). \]

### Step 3: Rewrite the Limit Expression

Now we can rewrite the limit expression:

\[ \lim_{x \to 1} \frac{(2x-3)(\sqrt{x}-1)}{2(x - 1)(x + \frac{3}{2})}. \]

### Step 4: Simplify the Numerator

Next, we notice that \( \sqrt{x} - 1 \) can be rewritten using the difference of squares:

\[ \sqrt{x} - 1 = \frac{x - 1}{\sqrt{x} + 1}. \]

Substituting this back into the limit gives:

\[ \lim_{x \to 1} \frac{(2x-3) \cdot \frac{x - 1}{\sqrt{x} + 1}}{2(x - 1)(x + \frac{3}{2})}. \]

### Step 5: Cancel Common Factors

Now we can cancel \( (x - 1) \) from the numerator and denominator:

\[ \lim_{x \to 1} \frac{(2x-3)}{2(\sqrt{x} + 1)(x + \frac{3}{2})}. \]

### Step 6: Substitute \( x = 1 \) Again

Now we substitute \( x = 1 \):

\[ 2(1) - 3 = -1, \]
\[ 2(\sqrt{1} + 1)(1 + \frac{3}{2}) = 2(1 + 1)(1 + 1.5) = 2 \cdot 2 \cdot 2.5 = 10. \]

### Final Result

Thus, the limit evaluates to:

\[ \frac{-1}{10} = -\frac{1}{10}. \]

So, the final answer is:

\[ \boxed{-\frac{1}{10}}. \]