Evaluate $\lim_{x \to \sqrt{2}} \dfrac{x^4 - 4}{x^2 + 3\sqrt{2}\,x - 8}$.

> [!TLDR] > This problem involves evaluating a limit of a rational function that initially results in an indeterminate $\frac{0}{0}$ form. The key idea is to factorize both the numerator and the denominator to cancel out the common factor responsible for the zero, then substitute the limit value. The final result is $\frac{8}{5}$. When evaluating limits of rational functions, our first step is always to try direct substitution. If this yields a finite number, that's our limit. However, if it results in an indeterminate form like $\frac{0}{0}$ or $\frac{\infty}{\infty}$, it means we need to simplify the expression further before substitution. The $\frac{0}{0}$ form tells us that $(x-a)$ is a factor of both the numerator and the denominator, where $a$ is the value $x$ approaches. Our goal is to find and cancel this common factor. 1. **Check for indeterminate form by direct substitution.** Let's substitute $x = \sqrt{2}$ into the numerator and the denominator. Numerator: $x^4 - 4 = (\sqrt{2})^4 - 4 = ((\sqrt{2})^2)^2 - 4 = (2)^2 - 4 = 4 - 4 = 0$ Denominator: $x^2 + 3\sqrt{2}\,x - 8 = (\sqrt{2})^2 + 3\sqrt{2}(\sqrt{2}) - 8 = 2 + 3(2) - 8 = 2 + 6 - 8 = 0$ Since we have the form $\frac{0}{0}$, direct substitution is not sufficient, and we must simplify the expression. This confirms that $(x-\sqrt{2})$ is a common factor in both the numerator and the denominator. 2. **Factorize the numerator.** The numerator is $x^4 - 4$. This is a difference of squares, $a^2 - b^2 = (a-b)(a+b)$, where $a=x^2$ and $b=2$. $x^4 - 4 = (x^2)^2 - 2^2 = (x^2 - 2)(x^2 + 2)$ The factor $(x^2 - 2)$ can be further factorized as a difference of squares, $x^2 - (\sqrt{2})^2 = (x-\sqrt{2})(x+\sqrt{2})$. So, the numerator becomes: $x^4 - 4 = (x-\sqrt{2})(x+\sqrt{2})(x^2+2)$ 3. **Factorize the denominator.** The denominator is a quadratic expression: $x^2 + 3\sqrt{2}\,x - 8$. Since we know that substituting $x=\sqrt{2}$ makes the denominator

Evaluate limx → √(2) dfracx⁴ - 4x² + 3√(2)x - 8. – Class 11 Mathematics

NCERT Exemplar · Q9

\lim_{x \to \sqrt{2}} \dfrac{x^4 - 4}{x^2 + 3\sqrt{2}\,x - 8}

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This problem involves evaluating a limit of a rational function that initially results in an indeterminate $\frac{0}{0}$ form. The key idea is to factorize both the numerator and the denominator to cancel out the common factor responsible for the zero, then substitute the limit value. The final result is $\frac{8}{5}$ .

When evaluating limits of rational functions, our first step is always to try direct substitution. If this yields a finite number, that's our limit. However, if it results in an indeterminate form like $\frac{0}{0}$ or $\frac{\infty}{\infty}$ , it means we need to simplify the expression further before substitution.

The $\frac{0}{0}$ form tells us that $(x-a)$ is a factor of both the numerator and the denominator, where $a$ is the value $x$ approaches. Our goal is to find and cancel this common factor.

Check for indeterminate form by direct substitution.

Let's substitute $x = \sqrt{2}$ into the numerator and the denominator.

Numerator:

$x^4 - 4 = (\sqrt{2})^4 - 4 = ((\sqrt{2})^2)^2 - 4 = (2)^2 - 4 = 4 - 4 = 0$

Denominator:

$x^2 + 3\sqrt{2}\,x - 8 = (\sqrt{2})^2 + 3\sqrt{2}(\sqrt{2}) - 8 = 2 + 3(2) - 8 = 2 + 6 - 8 = 0$

Since we have the form $\frac{0}{0}$ , direct substitution is not sufficient, and we must simplify the expression. This confirms that $(x-\sqrt{2})$ is a common factor in both the numerator and the denominator.
Factorize the numerator.

The numerator is $x^4 - 4$ . This is a difference of squares, $a^2 - b^2 = (a-b)(a+b)$ , where $a=x^2$ and $b=2$ .

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

The factor $(x^2 - 2)$ can be further factorized as a difference of squares, $x^2 - (\sqrt{2})^2 = (x-\sqrt{2})(x+\sqrt{2})$ .

So, the numerator becomes:

$x^4 - 4 = (x-\sqrt{2})(x+\sqrt{2})(x^2+2)$
Factorize the denominator.

The denominator is a quadratic expression: $x^2 + 3\sqrt{2}\,x - 8$ .

Since we know that substituting $x=\sqrt{2}$ makes the denominator zero, $(x-\sqrt{2})$ must be a factor. We can find the other factor by inspection or polynomial division.

Let the other factor be $(x+k)$ . Then:

$(x-\sqrt{2})(x+k) = x^2 + (k-\sqrt{2})x - k\sqrt{2}$

Comparing this to $x^2 + 3\sqrt{2}\,x - 8$ :
- Comparing the coefficient of $x$ : $k-\sqrt{2} = 3\sqrt{2} \implies k = 4\sqrt{2}$
- Comparing the constant term: $-k\sqrt{2} = -8 \implies -(4\sqrt{2})\sqrt{2} = -4(2) = -8$ . This matches.
So, the denominator factorizes as:

$x^2 + 3\sqrt{2}\,x - 8 = (x-\sqrt{2})(x+4\sqrt{2})$ …

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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.Evaluate $\lim_{x \to \sqrt{2}} \dfrac{x^4 - 4}{x^2 + 3\sqrt{2}\,x - 8}$ .

Concept understanding — Algebra of Limits

Q.Evaluate lim⁡x→2x4−4x2+32 x−8\lim_{x \to \sqrt{2}} \dfrac{x^4 - 4}{x^2 + 3\sqrt{2}\,x - 8}limx→2​​x2+32​x−8x4−4​.

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Q.Evaluate $\lim_{x \to \sqrt{2}} \dfrac{x^4 - 4}{x^2 + 3\sqrt{2}\,x - 8}$ .

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