PYQs Guides Notes Mocks Board Questions MHT-CET Maths Limits 137 MHT-CET Maths previous year questions (PYQs) from 68 papers, 2021 to 2026, with answers and worked solutions.
About this chapter Difficulty: 6 easy · 54 moderate · 77 hard. Most-asked subtopics: Continuity at a Point — Finding f(c) and the Parameter (34), Continuity of Piecewise Functions — Junction Conditions and Parameter Systems (24), Exponential, Logarithmic and 1^∞ Limits (21). Updated 13 October 2026. #1 · Limits — Existence, One-Sided Limits and Limits at Infinity
If lim x → ∞ [ x 3 + 1 x 2 + 1 − ( a x + b ) ] = 2 \displaystyle\lim_{x \to \infty}\left[\dfrac{x^3 + 1}{x^2 + 1} - (ax + b)\right] = 2 x → ∞ lim [ x 2 + 1 x 3 + 1 − ( a x + b ) ] = 2 then A a = 1 , b = 1 a = 1,\ b = 1 a = 1 , b = 1 B a = 1 , b = − 1 a = 1,\ b = -1 a = 1 , b = − 1 C a = 1 , b = − 2 a = 1,\ b = -2 a = 1 , b = − 2 D a = 1 , b = 2 a = 1,\ b = 2 a = 1 , b = 2 #2 · Limits — Existence, One-Sided Limits and Limits at Infinity
If f ( x ) = a x − 3 5 x + 7 f(x) = \dfrac{ax - 3}{5x + 7} f ( x ) = 5 x + 7 a x − 3 and lim x → 0 f ( x ) ⋅ lim x → ∞ f ( x ) = 3 \lim_{x \to 0}f(x) \cdot \lim_{x \to \infty}f(x) = 3 lim x → 0 f ( x ) ⋅ lim x → ∞ f ( x ) = 3 then a = … a = \ldots a = … #3 · Limits — Existence, One-Sided Limits and Limits at Infinity
The value of lim x → ∞ ( x 3 + 6 x 2 + 1 3 − x ) \lim_{x \to \infty}\left(\sqrt[3]{x^3 + 6x^2 + 1} - x\right) lim x → ∞ ( 3 x 3 + 6 x 2 + 1 − x ) is: #4 · Limits — Existence, One-Sided Limits and Limits at Infinity
The value of lim n → ∞ [ 1 1 − n 2 + 2 1 − n 2 + … + n 1 − n 2 ] 3 \lim_{n \to \infty}\left[\dfrac{1}{1 - n^2} + \dfrac{2}{1 - n^2} + \ldots + \dfrac{n}{1 - n^2}\right]^3 lim n → ∞ [ 1 − n 2 1 + 1 − n 2 2 + … + 1 − n 2 n ] 3 is MHT-CET · 2026 · 18 April Shift I · Q59 Hard Add #5 · Limits — Existence, One-Sided Limits and Limits at Infinity
If lim x → ∞ [ x 2 + x + 1 x + 1 − a x − b ] = 3 \lim_{x \to \infty}\left[\dfrac{x^2 + x + 1}{x + 1} - ax - b\right] = 3 lim x → ∞ [ x + 1 x 2 + x + 1 − a x − b ] = 3 , then a − b = a - b = a − b = ... Done a few? Here’s your next move
#6 · Limits — Existence, One-Sided Limits and Limits at Infinity
If lim x → ∞ ( 2 x − 1 ) 19 ⋅ ( 3 x + 2 ) 11 ( 6 x − 5 ) 30 = 2 a ⋅ 3 b \displaystyle\lim_{x \to \infty}\dfrac{(2x - 1)^{19}\cdot(3x + 2)^{11}}{(6x - 5)^{30}} = 2^a \cdot 3^b x → ∞ lim ( 6 x − 5 ) 30 ( 2 x − 1 ) 19 ⋅ ( 3 x + 2 ) 11 = 2 a ⋅ 3 b , then a + b = a + b = a + b = #7 · Limits — Existence, One-Sided Limits and Limits at Infinity
The value of lim n → ∞ ( n + 2 ) ! + ( n + 1 ) ! ( n + 2 ) ! − ( n + 1 ) ! \lim_{n \to \infty}\dfrac{(n + 2)! + (n + 1)!}{(n + 2)! - (n + 1)!} lim n → ∞ ( n + 2 )! − ( n + 1 )! ( n + 2 )! + ( n + 1 )! is ____ #8 · Limits — Existence, One-Sided Limits and Limits at Infinity
The value of lim x → 3 [ x ] − 3 x − 3 \lim_{x \to 3}\dfrac{[x] - 3}{x - 3} lim x → 3 x − 3 [ x ] − 3 , where [ ⋅ ] [\cdot] [ ⋅ ] denotes the greatest integer function, is..... #9 · Limits — Existence, One-Sided Limits and Limits at Infinity
lim x → ∞ ( 2 x + 1 ) 50 + ( 2 x + 2 ) 50 + ( 2 x + 3 ) 50 + . . . + ( 2 x + 100 ) 50 ( 2 x ) 50 + ( 10 ) 50 = \lim_{x\rightarrow \infty} \frac{(2x+ 1)^{50}+ (2x+ 2)^{50}+ (2x+ 3)^{50}+ ... + (2x+ 100)^{50}}{(2x)^{50}+ (10)^{50}}= lim x → ∞ ( 2 x ) 50 + ( 10 ) 50 ( 2 x + 1 ) 50 + ( 2 x + 2 ) 50 + ( 2 x + 3 ) 50 + ... + ( 2 x + 100 ) 50 = #10 · Limits — Existence, One-Sided Limits and Limits at Infinity
lim n → ∞ [ 1 1 − n 4 + 8 1 − n 4 + … … … … + n 3 1 − n 4 ] = \lim_{n\rightarrow \infty} \left\lbrack \frac{1}{1 -n^{4}}+\frac{8}{1 -n^{4}}+ \ldots\ldots\ldots\ldots +\frac{n^{3}}{1 -n^{4}} \right\rbrack= lim n → ∞ [ 1 − n 4 1 + 1 − n 4 8 + ………… + 1 − n 4 n 3 ] = #11 · Limits — Existence, One-Sided Limits and Limits at Infinity
lim x → ∞ e x 4 − 1 e x 4 + 1 = \lim_{x\rightarrow \infty} \frac{e^{x^{4}}- 1}{e^{x^{4}}+ 1}= lim x → ∞ e x 4 + 1 e x 4 − 1 = #12 · Limits — Existence, One-Sided Limits and Limits at Infinity
lim x → 0 ∣ x ∣ ∣ x ∣ + x 2 = \underset{x\rightarrow 0}{\lim} \frac{|x|}{|x| +x^{2}}= x → 0 lim ∣ x ∣ + x 2 ∣ x ∣ = #13 · Limits — Existence, One-Sided Limits and Limits at Infinity
If lim x → ∞ ( x 2 + x + 1 x + 1 − a x − b ) = 4 \lim_{x\to\infty}\left(\frac{x^2+x+1}{x+1} - ax - b\right) = 4 lim x → ∞ ( x + 1 x 2 + x + 1 − a x − b ) = 4 then A B a = 1 , b = − 4 a=1, b=-4 a = 1 , b = − 4 C a = 2 , b = − 3 a=2, b=-3 a = 2 , b = − 3 D MHT-CET · 2023 · 3rd May Shift 2 · Q138 Hard Add #14 · Limits — Existence, One-Sided Limits and Limits at Infinity
For each x ∈ R x\in\mathbb{R} x ∈ R , let [ x ] [x] [ x ] represent greatest integer function, then lim x → 0 − x ( [ x ] + ∣ x ∣ ) sin [ x ] ∣ x ∣ \lim_{x\to0^{-}}\frac{x([x]+|x|)\sin[x]}{|x|} lim x → 0 − ∣ x ∣ x ([ x ] + ∣ x ∣ ) s i n [ x ] is equal to #15 · Limits — Existence, One-Sided Limits and Limits at Infinity
The value of lim x → 0 x ∣ x ∣ + x 2 \lim_{x \to 0} \dfrac{x}{|x| + x^2} lim x → 0 ∣ x ∣ + x 2 x is #16 · Limits — Existence, One-Sided Limits and Limits at Infinity
lim x → ∞ x 3 { x 2 + 1 + x 4 − x 2 } = \lim_{x\to\infty}x^3\left\{\sqrt{x^2+\sqrt{1+x^4}}-x\sqrt2\right\}= lim x → ∞ x 3 { x 2 + 1 + x 4 − x 2 } = A B 1 4 2 \dfrac{1}{4\sqrt{2}} 4 2 1 C − 1 4 2 -\dfrac{1}{4\sqrt{2}} − 4 2 1 D − 1 2 -\dfrac{1}{\sqrt2} − 2 1 #17 · Limits — Existence, One-Sided Limits and Limits at Infinity
If l = lim x → 0 x ∣ x ∣ + x 2 l=\lim_{x\to0}\frac{x}{|x|+x^2} l = lim x → 0 ∣ x ∣ + x 2 x , then the value of l is: #18 · Algebraic Limits — Factorisation, Rationalisation and the xⁿ − aⁿ Form
If lim x → 1 x 3 + a x 2 + b x + c x 2 − 2 x + 1 = 2026 \lim_{x \to 1}\dfrac{x^3 + ax^2 + bx + c}{x^2 - 2x + 1} = 2026 lim x → 1 x 2 − 2 x + 1 x 3 + a x 2 + b x + c = 2026 then the value of a − c a - c a − c is... #19 · Algebraic Limits — Factorisation, Rationalisation and the xⁿ − aⁿ Form
If lim x → k x 3 − k 3 x 2 − k 2 = lim x → 0 1 − cos ( 2 x ) x sin x \lim_{x \to k}\dfrac{x^3 - k^3}{x^2 - k^2} = \lim_{x \to 0}\dfrac{1 - \cos(2x)}{x\sin x} lim x → k x 2 − k 2 x 3 − k 3 = lim x → 0 x sin x 1 − cos ( 2 x ) , then the value of k k k is … \ldots … #20 · Algebraic Limits — Factorisation, Rationalisation and the xⁿ − aⁿ Form
If lim x → 1 sin ( 3 x 2 − 4 x + 1 ) − x 2 + 1 2 x 3 − 7 x 2 + a x + b = − 2 \lim_{x \to 1}\dfrac{\sin(3x^2 - 4x + 1) - x^2 + 1}{2x^3 - 7x^2 + ax + b} = -2 lim x → 1 2 x 3 − 7 x 2 + a x + b sin ( 3 x 2 − 4 x + 1 ) − x 2 + 1 = − 2 , then the quadratic equation having roots a a a and b b b is A x 2 + 5 x − 24 = 0 x^2 + 5x - 24 = 0 x 2 + 5 x − 24 = 0 B x 2 − 5 x + 24 = 0 x^2 - 5x + 24 = 0 x 2 − 5 x + 24 = 0 C x 2 − 5 x − 24 = 0 x^2 - 5x - 24 = 0 x 2 − 5 x − 24 = 0 D x 2 + 5 x + 24 = 0 x^2 + 5x + 24 = 0 x 2 + 5 x + 24 = 0 #21 · Algebraic Limits — Factorisation, Rationalisation and the xⁿ − aⁿ Form
lim x → 1 [ x − 2 x 2 − x − 1 x 3 − 3 x 2 + 2 x ] = \lim_{x \to 1}\left[\dfrac{x - 2}{x^2 - x} - \dfrac{1}{x^3 - 3x^2 + 2x}\right] = lim x → 1 [ x 2 − x x − 2 − x 3 − 3 x 2 + 2 x 1 ] = #22 · Algebraic Limits — Factorisation, Rationalisation and the xⁿ − aⁿ Form
lim x → 2 x + 3 x 2 + 5 x 3 + 7 x 4 − 166 x − 2 = \lim_{x\rightarrow 2} \frac{x+ 3x^{2}+ 5x^{3}+ 7x^{4}- 166}{x- 2}= lim x → 2 x − 2 x + 3 x 2 + 5 x 3 + 7 x 4 − 166 = MHT-CET · 2025 · 19 April Shift I · Q108 Moderate Add #23 · Algebraic Limits — Factorisation, Rationalisation and the xⁿ − aⁿ Form
lim x → 3 ( 84 − x ) 1 4 − 3 x − 3 \underset{x\rightarrow 3}{\lim} \frac{(84 -x)^{\frac{1}{4}}- 3}{x- 3} x → 3 lim x − 3 ( 84 − x ) 4 1 − 3 is A − 1 108 \frac{- 1}{108} 108 − 1 B C D #24 · Algebraic Limits — Factorisation, Rationalisation and the xⁿ − aⁿ Form
lim y → 0 1 + 1 + y 4 − 2 y 4 = \lim_{y \to 0} \frac{\sqrt{1 + \sqrt{1+y^4}} - \sqrt{2}}{y^4} = lim y → 0 y 4 1 + 1 + y 4 − 2 = A B 1 2 2 \frac{1}{2\sqrt{2}} 2 2 1 C 1 4 2 \frac{1}{4\sqrt{2}} 4 2 1 D 1 2 2 ( 2 + 1 ) \frac{1}{2\sqrt{2}(\sqrt{2}+1)} 2 2 ( 2 + 1 ) 1 #25 · Algebraic Limits — Factorisation, Rationalisation and the xⁿ − aⁿ Form
If lim x → 1 x 2 − a x + b x − 1 = 7 \lim_{x\to1}\frac{x^2-ax+b}{x-1}=7 lim x → 1 x − 1 x 2 − a x + b = 7 , then a + b a+b a + b is equal to #26 · Algebraic Limits — Factorisation, Rationalisation and the xⁿ − aⁿ Form
Let α ( a ) \alpha(a) α ( a ) and β ( a ) \beta(a) β ( a ) be the roots of the equation ( 1 + a 3 − 1 ) x 2 + ( 1 + a − 1 ) x + ( 1 + a 6 − 1 ) = 0 (\sqrt[3]{1+a}-1)x^2+(\sqrt{1+a}-1)x+(\sqrt[6]{1+a}-1)=0 ( 3 1 + a − 1 ) x 2 + ( 1 + a − 1 ) x + ( 6 1 + a − 1 ) = 0 where a > − 1 a>-1 a > − 1 . Then lim a → 0 + α ( a ) \lim_{a\to0^+}\alpha(a) lim a → 0 + α ( a ) and lim a → 0 + β ( a ) \lim_{a\to0^+}\beta(a) lim a → 0 + β ( a ) respectively are A 1 and − 5 2 -\frac{5}{2} − 2 5 B − 1 -1 − 1 and − 1 2 -\frac{1}{2} − 2 1 C 2 2 2 and − 7 2 -\frac{7}{2} − 2 7 D 3 3 3 and − 9 2 -\frac{9}{2} − 2 9 #27 · Algebraic Limits — Factorisation, Rationalisation and the xⁿ − aⁿ Form
If f ( a ) = 2 , f ′ ( a ) = 1 , g ( a ) = − 1 , g ′ ( a ) = 2 f(a)=2, f'(a)=1, g(a)=-1, g'(a)=2 f ( a ) = 2 , f ′ ( a ) = 1 , g ( a ) = − 1 , g ′ ( a ) = 2 , then as x x x approaches a a a , g ( x ) f ( a ) − g ( a ) f ( x ) x − a \frac{g(x)f(a) - g(a)f(x)}{x-a} x − a g ( x ) f ( a ) − g ( a ) f ( x ) approaches #28 · Algebraic Limits — Factorisation, Rationalisation and the xⁿ − aⁿ Form
Let f ( x ) = 5 − ∣ x − 2 ∣ f(x) = 5 - |x-2| f ( x ) = 5 − ∣ x − 2∣ and g ( x ) = ∣ x + 1 ∣ g(x) = |x+1| g ( x ) = ∣ x + 1∣ , x ∈ R x\in\mathbb{R} x ∈ R . If f ( x ) f(x) f ( x ) attains maximum at α \alpha α and g ( x ) g(x) g ( x ) attains minimum at β \beta β , then lim x → − α β ( x − 1 ) ( x 2 − 5 x + 6 ) x 2 − 6 x + 8 \lim_{x\to -\alpha\beta}\frac{(x-1)(x^2-5x+6)}{x^2-6x+8} lim x → − α β x 2 − 6 x + 8 ( x − 1 ) ( x 2 − 5 x + 6 ) is equal to #29 · Algebraic Limits — Factorisation, Rationalisation and the xⁿ − aⁿ Form
lim x → 2 ( 1 x − 2 − 2 x x 3 − 3 x 2 + 2 x ) \lim_{x \to 2} \left(\frac{1}{x-2} - \frac{2x}{x^3-3x^2+2x}\right) lim x → 2 ( x − 2 1 − x 3 − 3 x 2 + 2 x 2 x ) is equal to #30 · Algebraic Limits — Factorisation, Rationalisation and the xⁿ − aⁿ Form
lim x → a a + 2 x − 3 x 3 a + x − 2 x = \displaystyle\lim_{x\to a}\frac{\sqrt{a+2x}-\sqrt{3x}}{\sqrt{3a+x}-2\sqrt{x}}= x → a lim 3 a + x − 2 x a + 2 x − 3 x = A 1 3 3 \dfrac{1}{3\sqrt{3}} 3 3 1 B C 2 3 3 \dfrac{2}{3\sqrt{3}} 3 3 2 D − 2 3 3 -\dfrac{2}{3\sqrt{3}} − 3 3 2 #31 · Algebraic Limits — Factorisation, Rationalisation and the xⁿ − aⁿ Form
Value of lim x → a a + 2 x − 3 x 3 a + x − 2 x \lim_{x\to a}\frac{\sqrt{a+2x}-\sqrt{3x}}{\sqrt{3a+x}-2\sqrt{x}} lim x → a 3 a + x − 2 x a + 2 x − 3 x is A 1 3 3 \frac{1}{3\sqrt{3}} 3 3 1 B C 2 3 3 \frac{2}{3\sqrt{3}} 3 3 2 D 4 3 3 \frac{4}{3\sqrt{3}} 3 3 4 #32 · Algebraic Limits — Factorisation, Rationalisation and the xⁿ − aⁿ Form
Evaluate lim x → 2 x 4 − 4 x 2 + 3 2 x − 8 \lim_{x \to \sqrt{2}} \dfrac{x^{4} - 4}{x^{2} + 3\sqrt{2}\,x - 8} lim x → 2 x 2 + 3 2 x − 8 x 4 − 4 . MHT-CET · 2022 · Shift 1 · Q134 Moderate Add #33 · Algebraic Limits — Factorisation, Rationalisation and the xⁿ − aⁿ Form
If lim x → 1 x 2 − a x + b x − 1 = 5 \lim_{x \to 1} \frac{x^2 - ax + b}{x - 1} = 5 lim x → 1 x − 1 x 2 − a x + b = 5 , then ( a + b ) (a+b) ( a + b ) is equal to #34 · Trigonometric Limits — sin x/x and the 1 − cos x Family
lim x → π 2 [ 1 − tan ( x 2 ) ] ( 1 − sin x ) [ 1 + tan ( x 2 ) ] ( π − 2 x ) 3 = \displaystyle\lim_{x \to \frac{\pi}{2}} \dfrac{[1 - \tan(\frac{x}{2})](1 - \sin x)}{[1 + \tan(\frac{x}{2})](\pi - 2x)^3} = x → 2 π lim [ 1 + tan ( 2 x )] ( π − 2 x ) 3 [ 1 − tan ( 2 x )] ( 1 − sin x ) = #35 · Trigonometric Limits — sin x/x and the 1 − cos x Family
The value of lim x → π 4 [ 4 − ( cos x + sin x ) 4 1 − sin 2 x ] \lim_{x \to \frac{\pi}{4}}\left[\dfrac{4 - (\cos x + \sin x)^4}{1 - \sin 2x}\right] lim x → 4 π [ 1 − sin 2 x 4 − ( cos x + sin x ) 4 ] is equal to #36 · Trigonometric Limits — sin x/x and the 1 − cos x Family
lim x → 0 1 − cos ( 1 − cos x ) x 4 = \lim_{x \to 0}\dfrac{1 - \cos(1 - \cos x)}{x^4} = lim x → 0 x 4 1 − cos ( 1 − cos x ) = #37 · Trigonometric Limits — sin x/x and the 1 − cos x Family
The value of lim x → 0 ( 8 x 8 ) [ 1 − cos x 2 2 − cos x 2 4 + cos x 2 2 ⋅ cos x 2 4 ] \lim_{x \to 0}\left(\dfrac{8}{x^8}\right)\left[1 - \cos\dfrac{x^2}{2} - \cos\dfrac{x^2}{4} + \cos\dfrac{x^2}{2} \cdot \cos\dfrac{x^2}{4}\right] lim x → 0 ( x 8 8 ) [ 1 − cos 2 x 2 − cos 4 x 2 + cos 2 x 2 ⋅ cos 4 x 2 ] is equal to ... MHT-CET · 2025 · 25 April Shift I · Q133 Hard Add #38 · Trigonometric Limits — sin x/x and the 1 − cos x Family
lim x → 5 2 − 2 cos ( x 2 − 12 x + 35 ) ( x − 5 ) = \lim_{x\rightarrow 5} \frac{\sqrt{2 - 2\cos\left( x^{2}- 12x+ 35 \right)}}{(x- 5)}= lim x → 5 ( x − 5 ) 2 − 2 c o s ( x 2 − 12 x + 35 ) = ...... . #39 · Trigonometric Limits — sin x/x and the 1 − cos x Family
lim x → π / 2 ( 1 − tan x 2 ) ( 1 − sin x ) ( 1 + tan x 2 ) ( π − 2 x ) 3 \lim_{x\to\pi/2}\frac{\left(1-\tan\frac{x}{2}\right)(1-\sin x)}{\left(1+\tan\frac{x}{2}\right)(\pi-2x)^{3}} lim x → π /2 ( 1 + t a n 2 x ) ( π − 2 x ) 3 ( 1 − t a n 2 x ) ( 1 − s i n x ) is #40 · Trigonometric Limits — sin x/x and the 1 − cos x Family
lim x → 0 x tan 2 x − 2 x tan x ( 1 − cos 2 x ) 2 \lim_{x\to 0}\frac{x\tan 2x-2x\tan x}{(1-\cos 2x)^2} lim x → 0 ( 1 − c o s 2 x ) 2 x t a n 2 x − 2 x t a n x is MHT-CET · 2023 · 9th May Shift 2 · Q131 Hard Add #41 · Trigonometric Limits — sin x/x and the 1 − cos x Family
lim x → π / 2 ( 1 − sin x ) ( 8 x 3 − π 3 ) cos x ( π − 2 x ) 4 \lim_{x\to\pi/2}\dfrac{(1-\sin x)(8x^3-\pi^3)\cos x}{(\pi-2x)^4} lim x → π /2 ( π − 2 x ) 4 ( 1 − sin x ) ( 8 x 3 − π 3 ) cos x is equal to A π 2 16 \frac{\pi^2}{16} 16 π 2 B 3 π 2 16 \frac{3\pi^2}{16} 16 3 π 2 C − 3 π 2 16 -\frac{3\pi^2}{16} − 16 3 π 2 D − π 2 16 -\frac{\pi^2}{16} − 16 π 2 #42 · Trigonometric Limits — sin x/x and the 1 − cos x Family
lim x → 0 ( 1 − cos 2 x ) ( 3 + cos x ) x tan 4 x \lim_{x\to0}\frac{(1-\cos2x)(3+\cos x)}{x\tan4x} lim x → 0 x t a n 4 x ( 1 − c o s 2 x ) ( 3 + c o s x ) has the value #43 · Trigonometric Limits — sin x/x and the 1 − cos x Family
lim x → 0 sin ( π cos 2 x ) x 2 \lim_{x\to0}\frac{\sin(\pi\cos^2 x)}{x^2} lim x → 0 x 2 s i n ( π c o s 2 x ) is equal to #44 · Trigonometric Limits — sin x/x and the 1 − cos x Family
lim x → 0 1 + x sin x − cos x tan 2 ( x / 2 ) = \lim_{x\to0}\dfrac{\sqrt{1+x\sin x}-\sqrt{\cos x}}{\tan^2(x/2)}= lim x → 0 tan 2 ( x /2 ) 1 + x sin x − cos x = #45 · Trigonometric Limits — sin x/x and the 1 − cos x Family
lim x → 0 x cot 4 x sin 2 x ⋅ cot 2 ( 2 x ) \lim_{x \to 0} \frac{x\cot 4x}{\sin^2 x \cdot \cot^2(2x)} lim x → 0 s i n 2 x ⋅ c o t 2 ( 2 x ) x c o t 4 x is equal to #46 · Trigonometric Limits — sin x/x and the 1 − cos x Family
lim x → 0 cos 7 x ∘ − cos 2 x ∘ x 2 \displaystyle\lim_{x\to0}\frac{\cos7x^{\circ}-\cos2x^{\circ}}{x^{2}} x → 0 lim x 2 cos 7 x ∘ − cos 2 x ∘ is A − 45 2 π 2 -\frac{45}{2}\pi^{2} − 2 45 π 2 B − 45 2 π -\frac{45}{2}\pi − 2 45 π C − π 2 1440 -\frac{\pi^{2}}{1440} − 1440 π 2 D − π 2 2880 -\frac{\pi^{2}}{2880} − 2880 π 2 #47 · Trigonometric Limits — sin x/x and the 1 − cos x Family
lim x → 0 ( 1 − cos 2 x ) ⋅ sin 5 x x 2 sin 3 x \lim_{x\to 0}\dfrac{(1-\cos 2x)\cdot\sin 5x}{x^2\sin 3x} lim x → 0 x 2 sin 3 x ( 1 − cos 2 x ) ⋅ sin 5 x is MHT-CET · 2024 · 9th May Shift 1 · Q138 Hard Add #48 · Trigonometric Limits — sin x/x and the 1 − cos x Family
lim x → π / 2 cot x − cos x ( π − 2 x ) 3 \lim_{x\to\pi/2}\frac{\cot x-\cos x}{(\pi-2x)^3} lim x → π /2 ( π − 2 x ) 3 c o t x − c o s x equals #49 · Exponential, Logarithmic and 1^∞ Limits
lim h → 0 log e ( 1 + 2 h ) − 2 log e ( 1 + h ) h 2 \displaystyle\lim_{h \to 0} \dfrac{\log_e(1 + 2h) - 2\log_e(1 + h)}{h^2} h → 0 lim h 2 log e ( 1 + 2 h ) − 2 log e ( 1 + h ) #50 · Exponential, Logarithmic and 1^∞ Limits
If lim x → 0 ( f ( x ) x 2 ) \lim_{x \to 0}\left(\dfrac{f(x)}{x^2}\right) lim x → 0 ( x 2 f ( x ) ) exists finitely and lim x → 0 ( 1 + x + f ( x ) x ) 1 / x = e 3 \lim_{x \to 0}\left(1 + x + \dfrac{f(x)}{x}\right)^{1/x} = e^3 lim x → 0 ( 1 + x + x f ( x ) ) 1/ x = e 3 , then ∫ f ( x ) ⋅ log x d x \int f(x) \cdot \log x\, dx ∫ f ( x ) ⋅ log x d x is equal to: A 2 9 x 3 ( 3 log x − 1 ) + c \dfrac{2}{9}x^3(3\log x - 1) + c 9 2 x 3 ( 3 log x − 1 ) + c B 1 9 x 3 ( 3 log x − 1 ) + c \dfrac{1}{9}x^3(3\log x - 1) + c 9 1 x 3 ( 3 log x − 1 ) + c C 2 9 x 3 ( 3 log x + 1 ) + c \dfrac{2}{9}x^3(3\log x + 1) + c 9 2 x 3 ( 3 log x + 1 ) + c D 2 9 x 3 ( 3 log x − 1 / 3 ) + c \dfrac{2}{9}x^3(3\log x - 1/3) + c 9 2 x 3 ( 3 log x − 1/3 ) + c Showing the first 50 of 137.
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