$\lim_{x \to 1} \dfrac{\left(\sqrt{x} - 1\right)(2x - 3)}{2x^2 + x - 3}$ is (A) $\dfrac{1}{10}$ (B) $\dfrac{-1}{10}$ (C) $1$ (D) None of these

limx → 1 dfrac(√(x) - 1)(2x - 3)2x² + x - 3 is (A) – Class 11 Mathematics

NCERT Exemplar · Q62

\lim_{x \to 1} \dfrac{\left(\sqrt{x} - 1\right)(2x - 3)}{2x^2 + x - 3}

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This limit is a $\frac{0}{0}$ form. Factor the denominator, cancel the common $(x-1)$ factor, then substitute $x=1$ to get the value $-\frac{1}{10}$ . The correct option is (B).

The key idea here is that when a limit gives $\frac{0}{0}$ , it means the numerator and denominator share a factor that becomes zero at the limit point. Our job is to find and cancel that factor — then the limit becomes straightforward.

Let’s look at the expression:

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

The numerator already has $\sqrt{x} - 1$ , which is zero at $x=1$ . The denominator $2x^2 + x - 3$ also becomes zero at $x=1$ (since $2+1-3=0$ ). So we have a $\frac{0}{0}$ form — cancellation is possible.

Factor the denominator $2x^2 + x - 3$ is a quadratic. We look for two numbers whose product is $2 \times (-3) = -6$ and sum is $1$ (the coefficient of $x$ ). Those numbers are $3$ and $-2$ . So:

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

Rewrite the limit

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

Connect $\sqrt{x} - 1$ to $x-1$ This is the crucial step. Recall the identity:

(\sqrt{x} - 1)(\sqrt{x} + 1) = x - 1

So $\sqrt{x} - 1 = \frac{x-1}{\sqrt{x} + 1}$ .

Substitute this into the numerator:

\frac{(\sqrt{x} - 1)(2x - 3)}{(2x+3)(x-1)} = \frac{\frac{x-1}{\sqrt{x} + 1} \cdot (2x - 3)}{(2x+3)(x-1)}

Cancel the common factor $x-1$ Since $x \neq 1$ in the limit (we only approach 1), $x-1 \neq 0$ and cancellation is valid: …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 13 on this concept.

CBSE 2025Set ANNUAL1 markMCQ
$\lim_{x \to 2} \left(\dfrac{x-2}{x^2-4}\right) =$
(a) 0
(b) 4
(c) $1/4$
(d) none of these
›Reveal solutionSolution

$\lim_{x\to2}\dfrac{x-2}{x^2-4}=\dfrac14$ .

Direct substitution gives $\dfrac00$ , an indeterminate form, so factor the denominator:

$x^2-4=(x-2)(x+2)$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$\lim_{\theta \to 0} \cos 2\theta =$
(a) 1
(b) -1
(c) 0
(d) 2
›Reveal solutionSolution

$\lim_{\theta\to0}\cos2\theta=1$ .

$\cos2\theta$ is continuous everywhere, so by direct substitution: …

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CBSE 2025Set ANNUAL1 markMCQ
$\displaystyle\lim_{x \to 4} \dfrac{4x + 3}{x - 2}$
(a) $\dfrac{19}{2}$
(b) $\dfrac{-19}{2}$
(c) $\dfrac{2}{19}$
(d) $\dfrac{-2}{19}$
›Reveal solutionSolution

Since the denominator $x - 2$ does not vanish at $x = 4$ , the limit is evaluated by direct substitution, giving $\dfrac{19}{2}$ .

We need $\displaystyle\lim_{x \to 4} \dfrac{4x+3}{x-2}$ .

…

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CBSE 2025Set ANNUAL1 mark
$\displaystyle\lim_{x \to 0} x \cdot \sec x$
›Reveal solutionSolution

Since $\sec x$ is continuous and finite at $x = 0$ (equal to 1), the limit of the product $x \sec x$ equals $0 \times 1 = 0$ .

…

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CBSE 2024Set ANNUAL1 markMCQ
$\displaystyle\lim_{x \to 2} \dfrac{x^3-2x^2}{x^2-5x+6} =$
(a) 4
(b) -4
(c) 0
(d) None of these
›Reveal solutionSolution

Both numerator and denominator vanish at $x=2$ (a $0/0$ form), so factor out the common $(x-2)$ term and cancel before substituting.

Numerator: $x^3 - 2x^2 = x^2(x-2)$

Denominator: $x^2 - 5x + 6 = (x-2)(x-3)$ (since $-2$ and $-3$ multiply to $6$ and add to $-5$ ).

So:

$\frac{x^3-2x^2}{x^2-5x+6} = \frac{x^2(x-2)}{(x-2)(x-3)} = \frac{x^2}{x-3} \quad (x e 2)$

…

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CBSE 2023Set ANNUAL1 markMCQ
$\displaystyle\lim_{x \to 2} \dfrac{x^3-2x^2}{x^2-5x+6} =$
(a) 4
(b) 0
(c) $-4$
(d) 2
›Reveal solutionSolution

Both numerator and denominator share a factor $(x-2)$ ; cancelling and substituting $x=2$ gives $-4$ .

Factor numerator and denominator:

$x^3-2x^2 = x^2(x-2)$

$x^2-5x+6 = (x-2)(x-3)$

So the expression simplifies (for $x\ne2$ ) to:

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

…

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CBSE 2022Set TERM11 markMCQ
$\displaystyle\lim_{x \to 2} \dfrac{x^2+x-6}{x-2} =$
(a) 5
(b) $-5$
(c) $\dfrac{0}{0}$
(d) $\dfrac{1}{2}$
›Reveal solutionSolution

Factor the quadratic numerator to cancel the common factor with the denominator.

…

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CBSE 2022Set ANNUAL1 markMCQ
$Q. Lim(x\to1) f(x) where f(x) = { x² - 1, x \leq 1 ; -x² - 1, x > 1 } is:$
(a) 0
(b) -2
(c) -1
(d) Does not exist
›Reveal solutionSolution

The left-hand limit is $0$ and the right-hand limit is $-2$ ; since they differ, the limit does not exist.

$f(x) = \begin{cases} x^2 - 1, & x \le 1 \\ -x^2 - 1, & x > 1 \end{cases}$

Left-hand limit (using $x^2-1$ since $x \le 1$ applies as $x \to 1^-$ ):

$\lim_{x\to 1^-} f(x) = 1^2 - 1 = 0$

Right-hand limit (using $-x^2-1$ since $x > 1$ applies as $x \to 1^+$ ): …

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CBSE 2020Set ANNUAL1 markMCQ
$\displaystyle\lim_{x \to 2} \frac{3x^2 - x - 10}{x^2 - 4} = ?$
(a) 0
(b) $\frac{1}{4}$
(c) $\frac{11}{4}$
(d) none of these
›Reveal solutionSolution

Both numerator and denominator vanish at $x=2$ (a $\frac00$ form); factor out the common $(x-2)$ term and cancel before substituting.

$\displaystyle\lim_{x\to2}\dfrac{3x^2-x-10}{x^2-4}$

Factor the numerator: $3x^2-x-10 = (x-2)(3x+5)$ (check: $(x-2)(3x+5)=3x^2+5x-6x-10=3x^2-x-10$ ✓)

…

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CBSE 2019Set ANNUAL1 markMCQ
$Q. The value of lim(x\to2) (x^3 - 2x^2) / (x^2 - 5x + 6) is:$
(a) 0
(b) 4
(c) -4
(d) None of these
›Reveal solutionSolution

Factor both numerator and denominator, cancel the common factor $(x-2)$ , then substitute $x=2$ .

Numerator: $x^3-2x^2 = x^2(x-2)$ .

Denominator: $x^2-5x+6 = (x-2)(x-3)$ .

$\dfrac{x^2(x-2)}{(x-2)(x-3)} = \dfrac{x^2}{x-3}$ for $x\ne2$ .

…

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CBSE 2019Set ANNUAL1 markMCQ
$Q. If f(x) = |x|/x for x \neq 0, and f(x) = 0 for x = 0, then lim(x\to0) f(x) is:$
(a) 0
(b) 1
(c) -1
(d) Does not exist
›Reveal solutionSolution

$f(x)=|x|/x$ equals $+1$ for $x>0$ and $-1$ for $x<0$ , so the left and right limits at $0$ disagree.

For $x>0$ : $f(x) = \dfrac{x}{x} = 1$ , so $\displaystyle\lim_{x\to0^+}f(x) = 1$ .

For $x<0$ : $f(x) = \dfrac{-x}{x} = -1$ , so $\displaystyle\lim_{x\to0^-}f(x) = -1$ .

…

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CBSE 2019Set annual1 mark
$f$
›Reveal solutionSolution

By the algebra of limits, the limit of a sum of two functions is the sum of their individual limits (when both exist).

One of the standard limit laws states that if $\displaystyle\lim_{x\to a} f(x)$ and $\displaystyle\lim_{x\to a} g(x)$ both exist, then the limit of their sum exists and equals the sum of the limits:

$\lim_{x \to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x)$

…

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Q. $\lim_{x \to 1} \dfrac{\left(\sqrt{x} - 1\right)(2x - 3)}{2x^2 + x - 3}$ is
(A) $\dfrac{1}{10}$
(B) $\dfrac{-1}{10}$
(C) $1$
(D) None of these

Concept understanding — Algebra of Limits

Q.lim⁡x→1(x−1)(2x−3)2x2+x−3\lim_{x \to 1} \dfrac{\left(\sqrt{x} - 1\right)(2x - 3)}{2x^2 + x - 3}limx→1​2x2+x−3(x​−1)(2x−3)​ is (A) 110\dfrac{1}{10}101​ (B) −110\dfrac{-1}{10}10−1​ (C) 111 (D) None of these

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Q. $\lim_{x \to 1} \dfrac{\left(\sqrt{x} - 1\right)(2x - 3)}{2x^2 + x - 3}$ is
(A) $\dfrac{1}{10}$
(B) $\dfrac{-1}{10}$
(C) $1$
(D) None of these

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