Limits54 questions

Limits — JEE Maths Practice Questions & Solutions

54 questions on Limits with full step-by-step solutions, including past-year (PYQ) problems. Free to practice.

hard
If variable line x(3+λ)+2y(2−λ)−(7−λ)=0x(3+\lambda)+2y(2-\lambda)-(7-\lambda)=0 always passes through a fixed point (a,b)(a,b) where λ\lambda is a parameter and λ=lim⁡x→(a−b)−[sin⁡x−2]+{cos⁡x}x−[x]−1\lambda=\displaystyle\lim_{x\to(a-b)^{-}}\dfrac{[\sin x-2]+\{\cos x\}}{x-[x]-1} where [y][y] and {y}\{y\} denotes greatest integer ≤\le and fractional part of yy respectively, then
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hard
The possible value(s) of kk for which lim⁡x→∞2x3−(tan⁡−1x)38πx3cot⁡−1(∣kx∣)+k2x6sin⁡(1x3)−3kx3=12\displaystyle\lim_{x\to\infty}\dfrac{2x^{3}-(\tan^{-1}x)^{3}}{\dfrac{8}{\pi}x^{3}\cot^{-1}(|kx|)+k^{2}x^{6}\sin\left(\dfrac{1}{x^{3}}\right)-3kx^{3}}=\dfrac{1}{2} is
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hard
Set of all values of xx such that lim⁡n→∞11+(4tan⁡−1(2πx)π)4n\displaystyle\lim_{n\to\infty}\dfrac{1}{1+\left(\dfrac{4\tan^{-1}(2\pi x)}{\pi}\right)^{4n}} is non-zero and finite number when n∈Nn\in\mathbb{N} is
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hard
Let a=min⁡[x2+2x+3,  x∈R]a=\min[x^{2}+2x+3,\;x\in\mathbb{R}] and b=lim⁡x→0sin⁡xcos⁡xex−e−xb=\displaystyle\lim_{x\to 0}\dfrac{\sin x\cos x}{e^{x}-e^{-x}}. Then the value of ∑r=0narbn−r\displaystyle\sum_{r=0}^{n}a^{r}b^{n-r} is
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medium
If α,β\alpha,\beta are roots of x2−1−cos⁡2θ x+θ=0x^{2}-\sqrt{1-\cos 2\theta}\,x+\theta=0 where 0<θ<π20<\theta<\dfrac{\pi}{2}, then
lim⁡θ→0−(1α+1β)= ?\displaystyle\lim_{\theta\to 0^{-}}\left(\dfrac{1}{\alpha}+\dfrac{1}{\beta}\right)=\,?
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medium
lim⁡x→∞x3sin⁡(1x)−2x21+3x2\displaystyle\lim_{x\to\infty}\dfrac{x^{3}\sin\left(\dfrac{1}{x}\right)-2x^{2}}{1+3x^{2}} is equal to
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medium
If lim⁡n→∞[(cos⁡kπ4)n−(cos⁡kπ6)n]=0\displaystyle\lim_{n\to\infty}\left[\left(\cos\dfrac{k\pi}{4}\right)^{n}-\left(\cos\dfrac{k\pi}{6}\right)^{n}\right]=0 (where kk is an integer), then which of the following statement(s) is(are) correct?
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medium
If lim⁡x→0((a−n)nx−tan⁡x)sin⁡nxx2=0\displaystyle\lim_{x\to 0}\dfrac{((a-n)nx-\tan x)\sin nx}{x^{2}}=0, where nn is a non-zero real number, then a= ?a=\,?
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medium
lim⁡x→0tan⁡([−π2]x2)−x2tan⁡([−π2])sin⁡2x\displaystyle\lim_{x\to 0}\dfrac{\tan\left([-\pi^{2}]x^{2}\right)-x^{2}\tan\left([-\pi^{2}]\right)}{\sin^{2}x}, where [⋅][\cdot] is GIF
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medium
Let f(x)={xtan⁡x,−∞<x<02,x=0ln⁡(1+2x)x,0<x<∞f(x)=\begin{cases}\dfrac{x}{\tan x}, & -\infty<x<0\\[4pt] 2, & x=0\\[4pt] \dfrac{\ln(1+2x)}{x}, & 0<x<\infty\end{cases} and g(x)={x+4,−∞<x<1x2−5x+11,1≤x<2x−3,2≤x≤∞g(x)=\begin{cases}x+4, & -\infty<x<1\\[4pt] x^{2}-5x+11, & 1\le x<2\\[4pt] x-3, & 2\le x\le\infty\end{cases} which of the following is(are) correct?
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medium
lim⁡x→0729x−243x−81x+9x+3x−1x3=K(log⁡3)3\displaystyle\lim_{x\to 0}\dfrac{729^{x}-243^{x}-81^{x}+9^{x}+3^{x}-1}{x^{3}}=K(\log 3)^{3}. Then K= ?K=\,?
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medium
If lim⁡x→0x(1+acos⁡x)−bsin⁡x(f(x))3=1\displaystyle\lim_{x\to 0}\dfrac{x(1+a\cos x)-b\sin x}{(f(x))^{3}}=1 and lim⁡x→0f(x)x=1\displaystyle\lim_{x\to 0}\dfrac{f(x)}{x}=1 then
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medium
If lim⁡x→0[1+x2+f(x)x]1/x2=e7\displaystyle\lim_{x\to 0}\left[1+x^{2}+\dfrac{f(x)}{x}\right]^{1/x^{2}}=e^{7}, then lim⁡x→0f(x)x3\displaystyle\lim_{x\to 0}\dfrac{f(x)}{x^{3}} is equal to.
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medium
Let f(x)=cos⁡−1(1−{x})sin⁡−1(1−{x})2{x}(1−{x})f(x)=\dfrac{\cos^{-1}(1-\{x\})\sin^{-1}(1-\{x\})}{\sqrt{2\{x\}}(1-\{x\})} where {x}\{x\} denotes the fractional part of xx, then
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medium
lim⁡x→0[∣sin⁡−1x∣x]\displaystyle\lim_{x\to 0}\left[\dfrac{|\sin^{-1}x|}{x}\right], where [⋅][\cdot] denotes greatest integer function, is
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medium
lim⁡x→0y→0y2+sin⁡xx2+sin⁡y2\displaystyle\lim_{\substack{x\to 0\\ y\to 0}}\dfrac{y^{2}+\sin x}{x^{2}+\sin y^{2}} when (x,y)→(0,0)(x,y)\to(0,0) along the curve x=y2x=y^{2} is
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medium
If α,β\alpha,\beta are roots of equation ax2+bx+c=0ax^{2}+bx+c=0 where 1<α<β1<\alpha<\beta. If lim⁡x→m∣ax2+bx+c∣ax2+bx+c=1\displaystyle\lim_{x\to m}\dfrac{|ax^{2}+bx+c|}{ax^{2}+bx+c}=1, then which of the following are true
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medium
Let f(x)={tan⁡2{x}x2−[x]2,x>01,x=0{x}cot⁡{x},x<0f(x)=\begin{cases}\dfrac{\tan^{2}\{x\}}{x^{2}-[x]^{2}}, & x>0\\[4pt] 1, & x=0\\[4pt] \sqrt{\{x\}\cot\{x\}}, & x<0\end{cases} where [x][x] is the step up function and {x}\{x\} is the fractional part function of xx, then
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medium
Let L=lim⁡x→1sin⁡(6cos⁡−1x)1−x2L=\displaystyle\lim_{x\to 1}\dfrac{\sin(6\cos^{-1}x)}{\sqrt{1-x^{2}}} and M=lim⁡x→11−cos⁡(6cos⁡−1x)1−x2M=\displaystyle\lim_{x\to 1}\dfrac{1-\cos(6\cos^{-1}x)}{1-x^{2}}. Which of the following is/are correct?
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medium
If lim⁡x→0eax−ex−xx2=b\displaystyle\lim_{x\to 0}\dfrac{e^{ax}-e^{x}-x}{x^{2}}=b (finite), then
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medium
ABCABC is an isosceles triangle inscribed in a circle of radius rr. If AB=ACAB=AC and hh is the altitude from AA to BCBC and PP be the perimeter of ABCABC then lim⁡h→0ΔP3\displaystyle\lim_{h\to 0}\dfrac{\Delta}{P^{3}} is equal to (where Δ\Delta is the area of the triangle)
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medium
lim⁡x→05sin⁡x−7sin⁡2x+3sin⁡3xx2sin⁡x\displaystyle\lim_{x\to 0}\dfrac{5\sin x-7\sin 2x+3\sin 3x}{x^{2}\sin x} is equal to
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medium
The value of the limit ∏n=2∞(1−1n2)\displaystyle\prod_{n=2}^{\infty}\left(1-\dfrac{1}{n^{2}}\right) is equal
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medium
The function f(x)=lim⁡n→∞x2n−1x2n+1f(x)=\displaystyle\lim_{n\to\infty}\dfrac{x^{2n}-1}{x^{2n}+1} is identical with the function.
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medium
The limiting value of f(x)=22−(cos⁡x+sin⁡x)31−sin⁡2xf(x)=\dfrac{2\sqrt{2}-(\cos x+\sin x)^{3}}{1-\sin 2x} when x→π4x\to\dfrac{\pi}{4} is
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medium
The value of lim⁡x→∞(2xn)1/ex−(3xn)1/exxn\displaystyle\lim_{x\to\infty}\dfrac{(2x^{n})^{1/e^{x}}-(3x^{n})^{1/e^{x}}}{x^{n}} (where n∈Nn\in\mathbb{N}) is
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medium
lim⁡n→∞12n+22(n−1)+32(n−2)+…+n2⋅113+23+33+…+n3\displaystyle\lim_{n\to\infty}\dfrac{1^{2}n+2^{2}(n-1)+3^{2}(n-2)+\ldots+n^{2}\cdot 1}{1^{3}+2^{3}+3^{3}+\ldots+n^{3}} equal
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medium
The value of lim⁡x→∞cot⁡−1(x−alog⁡ax)sec⁡−1(axlog⁡xa)\displaystyle\lim_{x\to\infty}\dfrac{\cot^{-1}(x^{-a}\log_{a}x)}{\sec^{-1}(a^{x}\log_{x}a)} (a>1a>1) is
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medium
Let f:(1,2)→Rf:(1,2)\to\mathbb{R} satisfies the inequality cos⁡(2x−4)−332<f(x)<x2∣4x−8∣x−2\dfrac{\cos(2x-4)-33}{2}<f(x)<\dfrac{x^{2}|4x-8|}{x-2}, ∀x∈(1,2)\forall x\in(1,2). Then lim⁡x→2−f(x)\displaystyle\lim_{x\to 2^{-}}f(x) is equal
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medium
Range of the function f(x)=[1ln⁡(x2+e)]+11+x2f(x)=\left[\dfrac{1}{\ln(x^{2}+e)}\right]+\dfrac{1}{\sqrt{1+x^{2}}}, [⋅][\cdot] denotes greatest integer function and e=lim⁡α→0(1+α)1/αe=\displaystyle\lim_{\alpha\to 0}(1+\alpha)^{1/\alpha}, is
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medium
If lim⁡x→0sin⁡(sin⁡x)−sin⁡xax5+bx3+c=−112\displaystyle\lim_{x\to 0}\dfrac{\sin(\sin x)-\sin x}{ax^{5}+bx^{3}+c}=-\dfrac{1}{12}, then
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medium
If f(0), f′(0), f′′(0)f(0),\,f'(0),\,f''(0) exist and are non-zero, and
lim⁡x→0af(2x)+2bf(x2)−3f(x)sin⁡−1(x2)=3, then\lim_{x\to 0}\dfrac{af(2x)+2bf\left(\dfrac{x}{2}\right)-3f(x)}{\sin^{-1}(x^{2})}=3, \text{ then}
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medium
If lim⁡x→0f(x)x2=a\displaystyle\lim_{x\to 0}\dfrac{f(x)}{x^{2}}=a and lim⁡x→0f(1−cos⁡x)g(x)sin⁡2x=b\displaystyle\lim_{x\to 0}\dfrac{f(1-\cos x)}{g(x)\sin^{2}x}=b where bb is not equal to zero then lim⁡x→0g(1−cos⁡2x)x4\displaystyle\lim_{x\to 0}\dfrac{g(1-\cos 2x)}{x^{4}} is
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easy
lim⁡x→∞x(x+1−x)\displaystyle\lim_{x\to\infty}\sqrt{x}\left(\sqrt{x+1}-\sqrt{x}\right) equal to
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easy
lim⁡x→∞∑r=110(x+r)2010(x1006+1)(2x1004+1) \displaystyle\lim_{x\to\infty}\dfrac{\sum_{r=1}^{10}(x+r)^{2010}}{(x^{1006}+1)(2x^{1004}+1)}\, is equal to
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easy
lim⁡x→0aex+bcos⁡x+ce−xe2x−2ex+1=4\displaystyle\lim_{x\to 0}\dfrac{ae^{x}+b\cos x+ce^{-x}}{e^{2x}-2e^{x}+1}=4. Then
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easy
The value of lim⁡x→1x1/3−x1/4x3−1\displaystyle \lim_{x\to 1}\dfrac{x^{1/3}-x^{1/4}}{x^{3}-1} is
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easy
lim⁡x→0ex−esin⁡x2(x−sin⁡x)= ?\displaystyle\lim_{x\to 0}\dfrac{e^{x}-e^{\sin x}}{2(x-\sin x)}=\,?
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easy
If lim⁡x→0g(x)\displaystyle\lim_{x\to 0}g(x) exists and lim⁡x→0g(x)⋅(e1/x−e−1/xe1/x+e−1/x)\displaystyle\lim_{x\to 0}g(x)\cdot\left(\dfrac{e^{1/x}-e^{-1/x}}{e^{1/x}+e^{-1/x}}\right) also exists, then g(x)g(x) is(are) equal to
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easy
If lim⁡x→∞4x(π4−tan⁡−1x+1x+2)=y2+4y+5\displaystyle\lim_{x\to\infty}4x\left(\dfrac{\pi}{4}-\tan^{-1}\dfrac{x+1}{x+2}\right)=y^{2}+4y+5, then yy can be equal to
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easy
Let f(x)=e{x}−{x}−1{x}2f(x)=\dfrac{e^{\{x\}}-\{x\}-1}{\{x\}^{2}} where {⋅}\{\cdot\} is fractional part of xx and [⋅][\cdot] is step function, then
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easy
Let L=lim⁡x→0a−a2−x2−x24x4L=\displaystyle\lim_{x\to 0}\dfrac{a-\sqrt{a^{2}-x^{2}}-\dfrac{x^{2}}{4}}{x^{4}}, a>0a>0. If LL is finite, then
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easy
lim⁡x→∞xp+xp−1+1xq+xq−2+2\displaystyle\lim_{x\to\infty}\dfrac{x^{p}+x^{p-1}+1}{x^{q}+x^{q-2}+2} (p>0,q>0p>0,q>0)
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easy
lim⁡h→0[1h8+h3−12h]\displaystyle\lim_{h\to 0}\left[\dfrac{1}{h\sqrt[3]{8+h}}-\dfrac{1}{2h}\right] is equal to
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easy
lim⁡x→2+{x}sin⁡(x−2)(x−2)2\displaystyle\lim_{x\to 2^{+}}\dfrac{\{x\}\sin(x-2)}{(x-2)^{2}} where {⋅}\{\cdot\} is fractional part function
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easy
If α,β∈(−π2,0)\alpha,\beta\in\left(-\dfrac{\pi}{2},0\right) such that (sin⁡α+sin⁡β)+sin⁡αsin⁡β=0(\sin\alpha+\sin\beta)+\dfrac{\sin\alpha}{\sin\beta}=0 and (sin⁡α+sin⁡β)⋅sin⁡αsin⁡β=−1(\sin\alpha+\sin\beta)\cdot\dfrac{\sin\alpha}{\sin\beta}=-1, and λ=lim⁡n→∞1+(2sin⁡α)2n(2sin⁡β)2n\lambda=\displaystyle\lim_{n\to\infty}\dfrac{1+(2\sin\alpha)^{2n}}{(2\sin\beta)^{2n}}, then
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easy
If α,β\alpha,\beta are the roots of ax2+bx+c=0ax^{2}+bx+c=0.
L1=lim⁡x→α1−cos⁡(ax2+bx+c)(x−α)2,L2=lim⁡x→β1−cos⁡(ax2+bx+c)(x−β)2L_{1}=\lim_{x\to\alpha}\dfrac{1-\cos(ax^{2}+bx+c)}{(x-\alpha)^{2}},\quad L_{2}=\lim_{x\to\beta}\dfrac{1-\cos(ax^{2}+bx+c)}{(x-\beta)^{2}}
L3=lim⁡(x−α)(x−β)→01−cos⁡(ax2+bx+c)(x−α)2(x−β)2. ThenL_{3}=\lim_{(x-\alpha)(x-\beta)\to 0}\dfrac{1-\cos(ax^{2}+bx+c)}{(x-\alpha)^{2}(x-\beta)^{2}}.\text{ Then}
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easy
lim⁡x→0 ⁣(sin⁡xx) ⁣ ⁣1/x2\displaystyle\lim_{x\to 0}\!\left(\dfrac{\sin x}{x}\right)^{\!\!1/x^{2}}
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easy
f(x)=13(f(x+1)+5f(x+2))f(x)=\dfrac{1}{3}\left(f(x+1)+\dfrac{5}{f(x+2)}\right) and f(x)>0,  ∀x∈Rf(x)>0,\;\forall x\in\mathbb{R}, then lim⁡x→∞f(x)\displaystyle\lim_{x\to\infty}f(x) is
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easy
lim⁡x→0([100sin⁡xx]+[100tan⁡xx])\displaystyle\lim_{x\to 0}\left(\left[\dfrac{100\sin x}{x}\right]+\left[\dfrac{100\tan x}{x}\right]\right), where [⋅][\cdot] is GIF
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easy
lim⁡x→π/2sin⁡xcos⁡−1[14(3sin⁡x−sin⁡3x)]\displaystyle\lim_{x\to\pi/2}\dfrac{\sin x}{\cos^{-1}\left[\dfrac{1}{4}(3\sin x-\sin 3x)\right]}, where [⋅][\cdot] is GIF
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easy
lim⁡x→∞cot⁡−1(x+1−x)sec⁡−1((2x+1x−1)x)\displaystyle\lim_{x\to\infty}\dfrac{\cot^{-1}\left(\sqrt{x+1}-\sqrt{x}\right)}{\sec^{-1}\left(\left(\dfrac{2x+1}{x-1}\right)^{x}\right)} is equal to
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easy
If f′(2)=6f'(2) = 6 and f′(1)=4f'(1) = 4, then lim⁡h→0f(2h+2+h2)−f(2)f(h−h2+1)−f(1)\displaystyle\lim_{h \to 0} \frac{f(2h+2+h^2)-f(2)}{f(h-h^2+1)-f(1)} is.
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easy
If f(x)=x⋅e∣x∣+[x]−2∣x∣+[x]f(x)=x\cdot\dfrac{e^{|x|+[x]}-2}{|x|+[x]} where [⋅][\cdot] is GIF, then
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