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: Try direct substitution.
Putting $x=1$ into $\frac{(2x-3)(\sqrt{x}-1)}{2x^2+x-3}$ gives $\frac{0}{0}$, so we must simplify before substituting.
Step 2: Factor the denominator.
$2x^2+x-3=2x^2+3x-2x-3=x(2x+3)-1(2x+3)=(2x+3)(x-1)$.
Step 3: Split the troublesome factor.
Note $x-1=(\sqrt{x}-1)(\sqrt{x}+1)$. So the denominator becomes $(2x+3)(\sqrt{x}-1)(\sqrt{x}+1)$.
Step 4: Cancel the common factor.
The $(\sqrt{x}-1)$ in the top cancels with the one in the bottom, leaving $\frac{2x-3}{(2x+3)(\sqrt{x}+1)}$.
Step 5: Substitute x=1 now.
Top: $2(1)-3=-1$. Bottom: $(2(1)+3)(\sqrt{1}+1)=(5)(2)=10$.
Step 6: Read the limit.
The limit is $\frac{-1}{10}=-\frac{1}{10}$, which is option (C).
\[ \boxed{\,-\tfrac{1}{10}\,} \]