Q.Suppose $f(x) = \begin{cases} a + bx, & x < 1 \\ 4, & x = 1 \\ b - ax, & x > 1 \end{cases}$. If $\displaystyle\lim_{x \to 1} f(x) = f(1)$, then what are the possible values of $a$ and $b$? OR Find the derivative of $f(x) = \cos x$ by first principle.

Showing the 12 most recent of 13 on this concept.

1. CBSE 2025Set ANNUAL1 markMCQ
   Q.$\lim_{x \to 2} \left(\dfrac{x-2}{x^2-4}\right) =$

   (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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2. CBSE 2025Set ANNUAL1 markMCQ
   Q.$\lim_{\theta \to 0} \cos 2\theta =$

   (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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3. CBSE 2025Set ANNUAL1 markMCQ
   Q.The value of $\displaystyle\lim_{x \to 4} \dfrac{4x + 3}{x - 2}$ will be:

   (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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4. CBSE 2025Set ANNUAL1 mark
   Q.Fill in the blank: The value of $\displaystyle\lim_{x \to 0} x \cdot \sec x$ is ____.

   ›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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5. CBSE 2024Set ANNUAL1 markMCQ
   Q.$\displaystyle\lim_{x \to 2} \dfrac{x^3-2x^2}{x^2-5x+6} =$

   (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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6. CBSE 2023Set ANNUAL1 markMCQ
   Q.$\displaystyle\lim_{x \to 2} \dfrac{x^3-2x^2}{x^2-5x+6} =$

   (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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7. CBSE 2022Set TERM11 markMCQ
   Q.$\displaystyle\lim_{x \to 2} \dfrac{x^2+x-6}{x-2} =$

   (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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8. CBSE 2022Set ANNUAL1 markMCQ
   Q.Lim(x→1) f(x) where f(x) = { x² - 1, x ≤ 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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9. CBSE 2020Set ANNUAL1 markMCQ
   Q.$\displaystyle\lim_{x \to 2} \frac{3x^2 - x - 10}{x^2 - 4} = ?$

   (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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10. CBSE 2019Set ANNUAL1 markMCQ
    Q.The value of lim(x→2) (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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11. CBSE 2019Set ANNUAL1 markMCQ
    Q.If f(x) = |x|/x for x ≠ 0, and f(x) = 0 for x = 0, then lim(x→0) 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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12. CBSE 2019Set annual1 mark
    Q.If '$f$' and '$g$' are two real functions such that both $\lim_{x \to a} f(x)$ and $\lim_{x \to a} g(x)$ exist, then $\lim_{x \to a} [f(x) + g(x)] =$ ............... . (Fill in the blank)

    ›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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