Q.$\lim_{r\to 1}\pi r^2$

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 $x=0$

   • $\cos 0 = 1$, $\sin 0 = 0$, $\sec^2 0 = 1$. So the limit is

   $\frac{2p \cdot 1 + 2 \cdot 0}{1 + 1} = \frac{2p}{2} = p.$

   4. Set equal to the given value

   We are told the original limit equals 1, so

   $p = 1.$

   ---

   Tip

   If you prefer series expansions:

   $\sin 2x \approx 2x$, $\cos 2x \approx 1 - 2x^2$, so numerator $\approx p(2x) + 1 - (1 - 2x^2) = 2px + 2x^2$.

   Denominator: $x + \tan x \approx x + x = 2x$.

   Then limit $\approx \frac{2px}{2x} = p$ as $x\to 0$. Same result.

   ---

   Watch out

   A common mistake is forgetting the factor 2 when differentiating $\sin 2x$ or $\cos 2x$. Always use the chain rule.

   ---

   ✓Final answer

   The correct option is (C).

   ANSWER: C

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:

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

   As $x \to \pi/2$, $\cos x \to 0$ and $\sin x \to 1$, so the denominator approaches $1+1 = 2$. Hence:

   $\frac{0}{2} = 0.$

   Tip

   A quick check: near $x = \pi/2$, let $x = \pi/2 - h$ with $h \to 0^+$. Then $\sin x = \cos h$ and $\cos x = \sin h$. The limit becomes $\frac{1-\cos h}{\sin h}$, which tends to $0$ because $1-\cos h \sim h^2/2$ and $\sin h \sim h$.

   Watch out

   A common mistake is to try L'Hôpital's Rule too early without simplifying — it works here too, but the algebraic manipulation is cleaner and avoids derivative errors.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

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

   $\frac{(1-\cos 2x)(3+\cos x)}{x\tan 4x} \sim \frac{8x^2}{4x^2} = 2.$

   Since all approximations are rigorous (they come from limits that exist), the original limit equals $2$.

   Tip

   A common shortcut: directly use $1-\cos\theta \sim \frac12\theta^2$ with $\theta=2x$ gives $1-\cos2x \sim 2x^2$, and $\tan4x\sim4x$. Then the whole fraction becomes $\frac{2x^2\cdot4}{x\cdot4x}=2$. Same result, fewer steps.

   Watch out

   Do not try to apply L’Hôpital’s rule without first simplifying — it works but is unnecessarily messy. The small‑angle method is faster and less error‑prone.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

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

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

   4. Take the limit. $\displaystyle\lim_{x\to3}\frac{x-3}{x^2+3x+9}=\frac{3-3}{9+9+9}=\frac{0}{27}=0.$

   Watch out

   Do not declare the limit infinite because each term diverges — the singularities cancel exactly, leaving a finite value $0$.

   ✓Final answer

   The correct option is (D) — the limit equals $0$.

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

   Dividing $x^3+3x^2-9x-2$ by $(x-2)$ (synthetic division with root $2$ on coefficients $1,3,-9,-2$):

   $x^3+3x^2-9x-2 = (x-2)(x^2+5x+1).$

   Step 4 — Cancel and evaluate the limit

   $\lim_{x\to 2} \frac{(x-2)(x^2+5x+1)}{(x-2)(x+2)(x-1)} = \lim_{x\to 2} \frac{x^2+5x+1}{(x+2)(x-1)} = \frac{4+10+1}{(4)(1)} = \frac{15}{4}.$

   ✓Final answer

   The correct option is (B) — $\frac{15}{4}$.

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

   $\frac{dy}{dx} = \frac{-1}{2\cos^{2}\!\left(\tfrac{\pi}{4}-\tfrac{x}{2}\right)} = -\frac{1}{2}\sec^{2}\!\left(\frac{\pi}{4} - \frac{x}{2}\right)$

   which is exactly statement (c) ✓. And statement (d), being its positive twin, is false ✗.

   5. Collect the verdicts. (a) ✓, (b) ✗, (c) ✓, (d) ✗ — statements (a) and (c) are two algebraically identical forms of the one derivative. So "Both a and c are correct".

   ✓Final answer

   The correct option is (C) Both a and c are correct — $\dfrac{dy}{dx} = \dfrac{-1}{1+\sin x} = -\dfrac{1}{2}\sec^{2}\!\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)$.

   ANSWER: C

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.

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

   Why this makes sense geometrically: for an odd function, the region on $[-1, 0]$ is the point-reflection of the region on $[0, 1]$ through the origin. The signed area below the axis on the left exactly cancels the signed area above it on the right, so the net value is zero.

   Note on the distractors: the options $\pi$, $\pi/2$ and $-\pi/2$ are irrelevant here — the limits are $\pm 1$, not $\pm\pi$, so no $\pi$ can enter the answer in the first place.

   ✓Final answer

   The correct option is (D) — $0$.

   ANSWER: D

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. Since the approximations are rigorous (they come from limits $\lim_{x\to0}\frac{1-\cos x}{x^2}=\frac12$ and $\lim_{x\to0}\frac{\sin x}{x}=1$), the original limit is exactly $\frac12$.

   Tip

   You can also do this without approximations by multiplying numerator and denominator by $1+\cos x$:

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

   As $x\to0$, $\frac{\sin x}{x}\to1$ and $\frac{1}{1+\cos x}\to\frac12$, giving the same result.

   Watch out

   A common mistake is to forget the sign change when substituting $\theta = \pi/2 - x$. Always check that $\cos\theta$ becomes $\sin x$ (not $-\sin x$), because $\cos(\pi/2 - x) = \sin x$ is positive for small $x$.

   ---

   ✓Final answer

   The correct option is (D).

   ANSWER: D

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 $x \to 1$

      Now plug $x=1$ into the simplified expression:

      Numerator: $2(1)-3 = -1$.

      Denominator: $(\sqrt{1}+1)(2(1)+3) = (1+1)(2+3) = 2 \cdot 5 = 10$.

      So the limit is $\displaystyle -\frac{1}{10}$.

   Tip

   Instead of using the conjugate, you could also set $t = \sqrt{x}$, so $x = t^2$, and as $x \to 1$, $t \to 1$. Then the limit becomes $\lim_{t \to 1} \frac{(t-1)(2t^2-3)}{2t^4 + t^2 - 3}$, factor $t-1$ from denominator, cancel, and get the same result — sometimes cleaner.

   Watch out

   A common mistake is to forget that $\sqrt{x} - 1$ is not directly a factor of $x-1$ — you must use the conjugate or substitution to reveal the hidden $(x-1)$ factor. Cancelling without that step leaves you stuck.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

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 $x = k$:

    $\frac{k^2 + k \cdot k + k^2}{k + k} = \frac{3k^2}{2k} = \frac{3k}{2},$

    provided $k \neq 0$ (and $k=0$ would make the original denominator zero in a different way, but we'll see it doesn't match).

    4. Set the two limits equal

    $4 = \frac{3k}{2} \quad \Rightarrow \quad k = \frac{8}{3}.$

    Tip

    You could also use L'Hôpital's rule directly on the right-hand limit:

    $\lim_{x \to k} \frac{x^3 - k^3}{x^2 - k^2} = \lim_{x \to k} \frac{3x^2}{2x} = \frac{3k^2}{2k} = \frac{3k}{2}$, same result.

    Watch out

    A common mistake is to forget that the right-hand limit is not simply the derivative of $\frac{x^3}{x^2}$ — you must treat the numerator and denominator separately because both approach zero.

    ---

    ✓Final answer

    The correct option is (D).

    ANSWER: D

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.

    Watch out

    A common mistake is to forget the chain rule factor of $-1$ from differentiating $(1-x)$, leading to the wrong answer $n$ (option A). Always check the inner derivative.

    Tip

    If you prefer algebra, expand $(1-x)^n = 1 - nx + \binom{n}{2}x^2 - \cdots$, then $\frac{(1-x)^n - 1}{x} = -n + \binom{n}{2}x - \cdots$, and as $x \to 0$ only $-n$ remains. Both methods agree.

    ✓Final answer

    The correct option is (C).

    ANSWER: C

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.

    Step 4 — Differentiate.

    $\frac{d}{dx}\left[\frac{2+x}{4}\right] = \frac{1}{4}.$

    This is a constant, which is exactly why the options are pure numbers with no $x$ in them — a strong confirmation that the intended route is "simplify, then differentiate."

    Sanity check: at $x=0$, $\cot\theta = 1 \Rightarrow \theta = 45^\circ \Rightarrow \cos^2\theta = \tfrac12$. Our formula gives $\tfrac{2+0}{4} = \tfrac12$. $\checkmark$

    ✓Final answer

    The correct option is (D) — $\dfrac{1}{4}$.

    ANSWER: D