Q.$\lim_{x\to 0}\dfrac{ax + b}{cx + 1}$

Showing the 12 most recent of 21 on this concept.

1. COMEDK 2026Set 2026-A1 markMCQ
   Q.If $\mathop {\lim }\limits_{x \to 0}\left(\frac{p \sin 2 x+1-\cos 2 x}{x+\tan x}\right)=1$ then the value of ' $p$ ' is (A) $2$ (B) $-1$ (C) $1$ (D) $\frac{1}{2}$

   ›Reveal solutionSolution

   The limit is of the form 0/0, so we apply L'Hôpital's rule (or series expansions) to find p such that the limit equals 1. The value of p is 1.

   We are given

   $\lim_{x\to 0} \frac{p\sin 2x + 1 - \cos 2x}{x + \tan x} = 1$

   and need to find $p$.

   ---

   Concept and intuition

   When $x \to 0$, both numerator and denominator approach 0 (since $\sin 0 = 0$, $\cos 0 = 1$, $\tan 0 = 0$). This is a $\frac{0}{0}$ indeterminate form. The standard tool is L'Hôpital's rule: differentiate numerator and denominator separately, then take the limit. Alternatively, we can use small-angle approximations (Taylor series) — both lead to the same condition on $p$.

   The key idea: the leading-order behavior of each term determines the limit. We'll match the limit to 1 by solving for $p$.

   ---

   Step-by-step solution

   1. Check the form at $x=0$

   • Numerator: $p\sin 0 + 1 - \cos 0 = 0 + 1 - 1 = 0$
   • Denominator: $0 + \tan 0 = 0$ So indeed $\frac{0}{0}$. L'Hôpital's rule applies.

   2. Apply L'Hôpital's rule

   Differentiate numerator and denominator:

   • Derivative of numerator:

     $\frac{d}{dx}[p\sin 2x] = p \cdot 2\cos 2x = 2p\cos 2x$

     $\frac{d}{dx}[1 - \cos 2x] = 2\sin 2x$

     So numerator derivative = $2p\cos 2x + 2\sin 2x$.

   • Derivative of denominator:

     $\frac{d}{dx}[x + \tan x] = 1 + \sec^2 x$.

   Thus the limit becomes

   $\lim_{x\to 0} \frac{2p\cos 2x + 2\sin 2x}{1 + \sec^2 x}.$

   3. Evaluate the new limit at $x=0$

   • $\cos 0 = 1$, $\sin 0 = 0$, $\sec^2 0 = 1$. …

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2. COMEDK 2026Set 2026-A1 markMCQ
   Q.

   [!FORMULA] $\mathop {\lim }\limits_{x \to {\pi \over 2}}\left(\frac{1-\sin x}{\cos x}\right) \text { is equal to }$

   (A) $1$ (B) $-1$ (C) $\frac{1}{2}$ (D) $0$

   ›Reveal solutionSolution

   The limit simplifies by factoring a difference of squares and using the identity $1-\sin^2 x = \cos^2 x$, yielding a value of $0$. The correct option is (D).

   We want

   $\lim_{x \to \pi/2} \frac{1-\sin x}{\cos x}.$

   If we plug in $x = \pi/2$ directly, we get $\frac{1-1}{0} = \frac{0}{0}$, an indeterminate form. So we need to manipulate the expression.

   Concept and intuition:

   The key is to notice that $1-\sin x$ and $\cos x$ are related through the Pythagorean identity $\sin^2 x + \cos^2 x = 1$. Rewriting $1-\sin x$ as a difference of squares can help cancel the $\cos x$ in the denominator. Specifically, $1-\sin^2 x = \cos^2 x$, and $1-\sin x$ is a factor of $1-\sin^2 x$.

   1. Multiply numerator and denominator by the conjugate The conjugate of $1-\sin x$ is $1+\sin x$. Multiply:

   $\frac{1-\sin x}{\cos x} \cdot \frac{1+\sin x}{1+\sin x} = \frac{1-\sin^2 x}{\cos x (1+\sin x)}.$

   2. Use the Pythagorean identity Since $1-\sin^2 x = \cos^2 x$, we have:

   $\frac{\cos^2 x}{\cos x (1+\sin x)} = \frac{\cos x}{1+\sin x},$

   provided $\cos x \neq 0$ (which is fine as we approach $\pi/2$, not equal to it).

   3. Evaluate the limit Now the expression is much simpler: $\lim_{x \to \pi/2} \frac{\cos x}{1+\sin x}.$ …

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3. COMEDK 2026Set 2026-M1 markMCQ
   Q.

   [!FORMULA] $\mathop {\lim }\limits_{x \to 0} \frac{(1-\cos 2 x)(3+\cos x)}{x \tan 4 x} \text { is equal to: }$

   (A) 3 (B) 2 (C) 4 (D) 1

   ›Reveal solutionSolution

   Use standard small‑angle approximations $\sin\theta\sim\theta$, $1-\cos\theta\sim\frac12\theta^2$, and $\tan\theta\sim\theta$ to reduce the limit to a simple constant; the limit equals $2$.

   Concept & Intuition

   When $x\to0$, all the trigonometric functions in the expression become small. The classic trick is to replace each factor by its leading-order Taylor expansion (or the equivalent small‑angle limit). This turns a messy fraction into a simple ratio of powers of $x$, from which the limit pops out directly. The key is to handle the numerator and denominator separately, using the identities

   $1-\cos 2x = 2\sin^2 x,\qquad \tan 4x \sim 4x,$

   and then apply $\sin x \sim x$.

   ---

   1. Rewrite the numerator using a double‑angle identity

   $1-\cos 2x = 2\sin^2 x.$

   So the numerator becomes

   $(1-\cos 2x)(3+\cos x) = 2\sin^2 x\,(3+\cos x).$

   2. Apply the small‑angle approximation for $\sin x$ As $x\to0$, $\sin x \sim x$. Hence $\sin^2 x \sim x^2$. Also $\cos x \to 1$, so $3+\cos x \to 4$. Therefore the numerator behaves like

   $2 \cdot x^2 \cdot 4 = 8x^2.$

   3. Simplify the denominator For small $x$, $\tan 4x \sim 4x$. Thus

   $x \tan 4x \sim x \cdot 4x = 4x^2.$

   4. Form the ratio and take the limit $$\frac{(1-\cos 2x)(3+\cos x)}{x\tan 4x} …$$

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4. COMEDK 2026Set 2026-M1 markMCQ
   Q.The value of $\mathop {\lim }\limits_{x \to 3}\left[\frac{1}{x-3}+\frac{9 x}{27-x^3}\right]$ is: (A) $\frac{1}{2}$ (B) 1 (C) 2 (D) 0

   ›Reveal solutionSolution

   The expression is an $\infty-\infty$ form; combining the fractions and factoring the difference of cubes cancels the singular $(x-3)$ factor and leaves $\frac{x-3}{x^2+3x+9}$, whose limit as $x\to3$ is $0$. The correct option is (D).

   Concept

   Substituting $x=3$ makes both $\frac{1}{x-3}$ and $\frac{9x}{27-x^3}$ blow up, giving an indeterminate $\infty-\infty$. Combining them into one fraction and factoring $27-x^3$ with $a^3-b^3=(a-b)(a^2+ab+b^2)$ exposes a cancelling $(x-3)$ factor, revealing the true finite limit.

   Solution

   1. Factor the cubic. $27-x^3=-(x-3)(x^2+3x+9)$.
   2. Combine.

   $\frac{1}{x-3}-\frac{9x}{(x-3)(x^2+3x+9)}=\frac{(x^2+3x+9)-9x}{(x-3)(x^2+3x+9)}.$

   3. Simplify the numerator. $x^2+3x+9-9x=x^2-6x+9=(x-3)^2$, so …

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5. KCET 2026Set UNKNOWN1 markMCQ
   Q.The value of $\lim_{x\to 2} \frac{x^3+3x^2-9x-2}{x^3-x^2-4x+4}$ is ________ (A) 3 (B) $\frac{15}{4}$ (C) $\frac{15}{2}$ (D) $\frac{15}{13}$

   ›Reveal solutionSolution

   Both the numerator and denominator equal $0$ at $x=2$, giving a $\frac{0}{0}$ indeterminate form — factor out $(x-2)$ from each and cancel before evaluating the limit.

   Step 1 — Confirm the indeterminate form

   At $x=2$: numerator $= 8+12-18-2 = 0$; denominator $= 8-4-8+4 = 0$. So $(x-2)$ is a common factor of both.

   Step 2 — Factor the denominator

   Dividing $x^3-x^2-4x+4$ by $(x-2)$ (synthetic division with root $2$ on coefficients $1,-1,-4,4$):

   $x^3-x^2-4x+4 = (x-2)(x^2+x-2) = (x-2)(x+2)(x-1).$

   Step 3 — Factor the numerator …

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6. KCET 2025Set A-11 markMCQ
   Q.If $y = \dfrac{\cos x}{1+\sin x}$, then

   (a) $\dfrac{dy}{dx} = \dfrac{-1}{1+\sin x}$

   (b) $\dfrac{dy}{dx} = \dfrac{1}{1+\sin x}$

   (c) $\dfrac{dy}{dx} = -\dfrac{1}{2}\sec^{2}\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)$

   (d) $\dfrac{dy}{dx} = \dfrac{1}{2}\sec^{2}\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)$ (A) Only b is correct (B) Only a is correct (C) Both a and c are correct (D) Both b and d are correct

   ›Reveal solutionSolution

   Differentiate by the quotient rule to get $-\dfrac{1}{1+\sin x}$ (statement a), then use the half-angle identity $1+\sin x = 2\cos^{2}\!\left(\tfrac{\pi}{4}-\tfrac{x}{2}\right)$ to see that this is identical to statement (c).

   1. Differentiate using the quotient rule. With $u = \cos x$, $v = 1 + \sin x$:

   $\frac{dy}{dx} = \frac{v\,u' - u\,v'}{v^{2}} = \frac{(1+\sin x)(-\sin x) - (\cos x)(\cos x)}{(1+\sin x)^{2}}$

   2. Simplify the numerator.

   $(1+\sin x)(-\sin x) - \cos^{2}x = -\sin x - \sin^{2}x - \cos^{2}x$

   Use the Pythagorean identity $\sin^2 x + \cos^2 x = 1$:

   $= -\sin x - 1 = -(1 + \sin x)$

   3. Cancel.

   $\frac{dy}{dx} = \frac{-(1+\sin x)}{(1+\sin x)^{2}} = \boxed{\frac{-1}{1+\sin x}}$

   This is exactly statement (a) ✓, and it immediately makes statement (b) (the same thing with a $+$ sign) false ✗.

   4. Now test statement (c) — the half-angle rewrite. Convert $1 + \sin x$ into a perfect square. Since $\sin x = \cos\!\left(\tfrac{\pi}{2} - x\right)$ and $1 + \cos\theta = 2\cos^{2}\!\left(\tfrac{\theta}{2}\right)$, put $\theta = \tfrac{\pi}{2} - x$:

   $1 + \sin x = 1 + \cos\!\left(\frac{\pi}{2}-x\right) = 2\cos^{2}\!\left(\frac{\pi}{4} - \frac{x}{2}\right)$

   Therefore …

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7. KCET 2025Set A-11 markMCQ
   Q.The value of $\int_{-1}^{1} \sin^5 x \cos^4 x \, dx$ is (A) $-\pi/2$ (B) $\pi$ (C) $\pi/2$ (D) $0$

   ›Reveal solutionSolution

   The integrand is an odd function integrated over the symmetric interval $[-1,1]$, so the integral vanishes.

   Step 1 — Recall the symmetry property of definite integrals.

   $\int_{-a}^{a} f(x)\,dx = \begin{cases} 2\displaystyle\int_{0}^{a} f(x)\,dx, & \text{if } f \text{ is even } (f(-x)=f(x))\\[2mm] 0, & \text{if } f \text{ is odd } (f(-x)=-f(x))\end{cases}$

   So the whole problem reduces to testing the parity of the integrand — no antiderivative is needed.

   Step 2 — Test the parity of $f(x) = \sin^5 x \cos^4 x$.

   Use the basic parities $\sin(-x) = -\sin x$ (odd) and $\cos(-x) = \cos x$ (even):

   $f(-x) = \sin^5(-x)\,\cos^4(-x) = (-\sin x)^5 (\cos x)^4 = -\sin^5 x \cos^4 x = -f(x)$

   The minus sign survives because the sine is raised to an odd power ($5$), while the cosine's even power ($4$) kills any sign change. So $f$ is odd.

   Step 3 — Check the limits.

   The interval is $[-1, 1]$, symmetric about $x = 0$ with $a = 1$. Both conditions of the odd-function rule hold.

   Step 4 — Conclude.

   $\int_{-1}^{1}\sin^5 x\cos^4 x\,dx = 0$ …

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8. COMEDK 2025Set 2025-A1 markMCQ
   Q.$\lim _{\theta \rightarrow \frac{\pi}{2}} \frac{1-\sin \theta}{\left(\frac{\pi}{2}-\theta\right) \cos \theta}$ is equal to : (A) $-\frac{1}{2}$ (B) $-1$ (C) $1$ (D) $\frac{1}{2}$

   ›Reveal solutionSolution

   This limit is a classic 0/0 indeterminate form that simplifies using the substitution $x = \frac{\pi}{2} - \theta$ and the small‑angle approximation $\sin x \approx x$. The limit equals $\frac12$, so the correct option is (D).

   ---

   We want

   $\lim_{\theta \to \frac{\pi}{2}} \frac{1 - \sin\theta}{\left(\frac{\pi}{2} - \theta\right) \cos\theta}.$

   Direct substitution gives $\frac{0}{0}$, so we need a clever rewrite.

   Concept & Intuition

   When a limit involves $\theta$ near $\pi/2$, it’s often easier to shift the variable so the “trouble spot” moves to 0. Let

   $x = \frac{\pi}{2} - \theta \quad \Longrightarrow \quad \theta = \frac{\pi}{2} - x.$

   As $\theta \to \pi/2$, we have $x \to 0$. This substitution turns the trigonometric functions into expressions in $x$ that we can expand using known small‑angle approximations.

   ---

   Step‑by‑step

   1. Rewrite $\sin\theta$ and $\cos\theta$ in terms of $x$.

   $\sin\theta = \sin\!\left(\frac{\pi}{2} - x\right) = \cos x, \qquad \cos\theta = \cos\!\left(\frac{\pi}{2} - x\right) = \sin x.$

   2. Substitute everything into the limit. The numerator becomes

   $1 - \sin\theta = 1 - \cos x.$

   The denominator becomes

   $\left(\frac{\pi}{2} - \theta\right) \cos\theta = x \cdot \sin x.$

   So the limit transforms to

   $\lim_{x \to 0} \frac{1 - \cos x}{x \sin x}.$

   3. Recognise the standard small‑angle forms. For $x \to 0$:

   $1 - \cos x \sim \frac{x^2}{2}, \qquad \sin x \sim x.$

   These are the first terms of the Taylor series; they are exact in the limit.

   4. Apply the approximations.

   $\frac{1 - \cos x}{x \sin x} \sim \frac{\frac{x^2}{2}}{x \cdot x} = \frac{\frac{x^2}{2}}{x^2} = \frac12.$

   The $x^2$ cancels, leaving a constant.

   5. Conclude the limit. …

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9. COMEDK 2025Set 2025-A1 markMCQ
   Q.$\lim _{x \rightarrow 1} \frac{(\sqrt{x}-1)(2 x-3)}{2 x^2+x-3}$ is (A) $\frac{1}{10}$ (B) 0 (C) 1 (D) $-\frac{1}{10}$

   ›Reveal solutionSolution

   The limit is a $\frac{0}{0}$ indeterminate form, so we factor and cancel the common $(x-1)$ term; after simplification, the limit evaluates to $-\frac{1}{10}$, which corresponds to option (D).

   Concept & Intuition

   When you plug $x=1$ directly into the expression, both numerator and denominator become zero — that’s the classic $\frac{0}{0}$ indeterminate form. The trick is to “remove” the common factor causing the zero. Here, the numerator has a $\sqrt{x} - 1$ factor (which is zero at $x=1$), and the denominator is a quadratic that also vanishes at $x=1$. By factoring both, we can cancel the problematic $(x-1)$ (or its equivalent) and then safely substitute.

   1. Check direct substitution

      Let $f(x) = \frac{(\sqrt{x}-1)(2x-3)}{2x^2 + x - 3}$.

      At $x=1$:

      $\sqrt{1} - 1 = 0$, so numerator $= 0 \cdot (2\cdot1 - 3) = 0$.

      Denominator: $2(1)^2 + 1 - 3 = 2 + 1 - 3 = 0$.

      So we have $\frac{0}{0}$ — need to simplify.

   2. Factor the denominator

      $2x^2 + x - 3$ is a quadratic. Look for factors:

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

      Check: $(2x+3)(x-1) = 2x^2 -2x + 3x -3 = 2x^2 + x -3$ ✓.

      So denominator $= (2x+3)(x-1)$.

   3. Rewrite the numerator’s $\sqrt{x} - 1$ in terms of $x-1$

      Recall the identity: $\sqrt{x} - 1 = \frac{x-1}{\sqrt{x} + 1}$ (multiply numerator and denominator by conjugate).

      Thus numerator becomes:

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

   4. Cancel the common $(x-1)$ factor

      The whole expression is now:

      $\displaystyle \frac{\frac{x-1}{\sqrt{x}+1} \cdot (2x-3)}{(2x+3)(x-1)}$.

      Cancel $(x-1)$ (valid for $x \neq 1$, which is fine since we’re taking a limit):

      $\displaystyle \frac{2x-3}{(\sqrt{x}+1)(2x+3)}$.

   5. Take the limit as $x \to 1$ …

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10. COMEDK 2025Set 2025-E1 markMCQ
    Q.If $\lim\limits_{x \rightarrow 1} \frac{x^4-1}{x-1}=\lim\limits_{x \rightarrow k} \frac{x^3-k^3}{x^2-k^2}$, then the value of K is : (A) $\frac{3}{4}$ (B) $\frac{4}{3}$ (C) $8$ (D) $\frac{8}{3}$

    ›Reveal solutionSolution

    The problem equates two limits that are both derivatives at a point. The left side is the derivative of $x^4$ at $x=1$, giving $4$. The right side simplifies to a derivative expression in $k$, leading to $k = \frac{8}{3}$, so the correct option is (D).

    We are given:

    $\lim_{x \to 1} \frac{x^4 - 1}{x - 1} = \lim_{x \to k} \frac{x^3 - k^3}{x^2 - k^2}$

    and asked to find $k$.

    ---

    Concept and Intuition

    Each limit is of the form $\frac{f(x) - f(a)}{x - a}$ as $x \to a$, which is exactly the definition of the derivative $f'(a)$.

    So the left side is $f'(1)$ for $f(x) = x^4$, and the right side is a derivative-like expression but with numerator and denominator both zero at $x = k$. That means we can factor and cancel, or recognize it as a ratio of derivatives via L'Hôpital's rule.

    ---

    Step-by-step solution

    1. Evaluate the left-hand limit

    $\lim_{x \to 1} \frac{x^4 - 1}{x - 1}$

    This is the derivative of $x^4$ at $x = 1$:

    $\frac{d}{dx}(x^4) = 4x^3 \quad \Rightarrow \quad 4(1)^3 = 4.$

    So the left side equals $4$.

    2. Simplify the right-hand limit

    $\lim_{x \to k} \frac{x^3 - k^3}{x^2 - k^2}$

    Both numerator and denominator vanish at $x = k$, so we can factor:

    $x^3 - k^3 = (x - k)(x^2 + xk + k^2), \quad x^2 - k^2 = (x - k)(x + k).$

    Cancel $(x - k)$ (valid for $x \neq k$, and the limit doesn't care about the point itself):

    $\lim_{x \to k} \frac{x^2 + xk + k^2}{x + k}.$

    3. Evaluate the simplified limit Substitute $x = k$: $\frac{k^2 + k \cdot k + k^2}{k + k} = \frac{3k^2}{2k} = \frac{3k}{2},$ …

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11. COMEDK 2025Set 2025-M1 markMCQ
    Q.The value of $\lim _{x \rightarrow 0} \frac{(1-x)^n-1}{x}=$ (A) n (B) 0 (C) $-$n (D) 1

    ›Reveal solutionSolution

    This limit is the definition of the derivative of $(1-x)^n$ at $x=0$, which gives $-n$ as the answer.

    The core idea is recognizing that the expression $\frac{(1-x)^n - 1}{x}$ looks exactly like the difference quotient for the function $f(x) = (1-x)^n$ at $x=0$.

    When we compute $\lim_{x \to 0} \frac{f(x) - f(0)}{x}$, we are finding $f'(0)$.

    So instead of algebraic manipulation, we can differentiate directly — faster and less error-prone.

    1. Identify the function and the point.

       Let $f(x) = (1-x)^n$. Then $f(0) = 1^n = 1$.

       The given limit is $\lim_{x \to 0} \frac{f(x) - f(0)}{x - 0}$, which is precisely $f'(0)$.

    2. Differentiate $f(x)$.

       Using the chain rule:

       $f'(x) = n(1-x)^{n-1} \cdot (-1) = -n(1-x)^{n-1}$.

    3. Evaluate at $x=0$.

       $f'(0) = -n(1-0)^{n-1} = -n \cdot 1 = -n$.

    4. Therefore the limit equals $-n$.

       No need to expand $(1-x)^n$ or apply L'Hôpital's rule — the derivative interpretation gives the answer instantly. …

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12. KCET 2024Set A-11 markMCQ
    Q.$\dfrac{d}{dx} \left[ \cos^2 \cot^{-1} \sqrt{\dfrac{2+x}{2-x}} \right]$ is (A) $-\dfrac{3}{4}$ (B) $-\dfrac{1}{2}$ (C) $\dfrac{1}{2}$ (D) $\dfrac{1}{4}$

    ›Reveal solutionSolution

    Simplify before differentiating: the expression $\cos^2\!\left(\cot^{-1}t\right)$ collapses to $\frac{t^2}{1+t^2}$, which here reduces to the linear function $\frac{2+x}{4}$.

    Step 1 — Name the inverse-trig angle.

    Let

    $\theta = \cot^{-1}\sqrt{\frac{2+x}{2-x}} \quad\Longrightarrow\quad \cot\theta = \sqrt{\frac{2+x}{2-x}}, \qquad \cot^{2}\theta = \frac{2+x}{2-x}.$

    The expression to differentiate is simply $\cos^2\theta$.

    Step 2 — Express $\cos^2\theta$ in terms of $\cot^2\theta$.

    This is the crucial identity. Since $\csc^2\theta = 1 + \cot^2\theta$ and $\cos^2\theta = \dfrac{\cot^2\theta}{\csc^2\theta}$ (because $\cos^2 = \frac{\cos^2}{\sin^2}\cdot\sin^2 = \cot^2\theta\cdot\sin^2\theta$):

    $\cos^{2}\theta = \frac{\cot^{2}\theta}{1+\cot^{2}\theta}.$

    Step 3 — Substitute and simplify.

    Denominator first:

    $1 + \cot^{2}\theta = 1 + \frac{2+x}{2-x} = \frac{(2-x)+(2+x)}{2-x} = \frac{4}{2-x}.$

    Therefore

    $\cos^{2}\theta = \frac{\dfrac{2+x}{2-x}}{\dfrac{4}{2-x}} = \frac{2+x}{2-x}\cdot\frac{2-x}{4} = \frac{2+x}{4}.$

    The messy square-root and inverse-cotangent have vanished entirely — the bracket is just a straight line. …

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