Question

$\lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{2 x^2+x-3}=$

• A. $\frac{1}{5}$
• B. $\frac{1}{10}$
• C. $-\frac{1}{10}$
• D. $-\frac{1}{5}$

Hint:

Whenever you see a term like $(\sqrt{x}-1)$ in a limit involving polynomials, try to factorize the polynomial as a difference of squares or recognize $(x-1)$ as $(\sqrt{x}-1)(\sqrt{x}+1)$.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are asked to evaluate the limit of a rational function as $x$ approaches 1. Direct substitution results in $\frac{0}{0}$, indicating an indeterminate form.

Step 2: Key Formula or Approach:
Factorize the quadratic denominator $2x^2 + x - 3$ and simplify the expression to resolve the indeterminacy.

Step 3: Detailed Explanation:
The denominator $2x^2 + x - 3$ can be factored: $2x^2 + 3x - 2x - 3 = x(2x + 3) - 1(2x + 3) = (2x + 3)(x - 1)$.
Recall that $(x-1) = (\sqrt{x}-1)(\sqrt{x}+1)$.
So, the denominator is $(2x + 3)(\sqrt{x} - 1)(\sqrt{x} + 1)$.
Substituting this into the limit:
$\lim_{x \to 1} \frac{(2x-3)(\sqrt{x}-1)}{(2x+3)(\sqrt{x}-1)(\sqrt{x}+1)} = \lim_{x \to 1} \frac{2x-3}{(2x+3)(\sqrt{x}+1)}$
Evaluating the limit at $x=1$:
$\frac{2(1)-3}{(2(1)+3)(\sqrt{1}+1)} = \frac{-1}{(5)(2)} = -\frac{1}{10}$.

Step 4: Final Answer:
The value of the limit is $-\frac{1}{10}$, which corresponds to option (C).