Limits: JEE Questions & Solutions

Let f: R → (0, ∞) be a twice differentiable function such that f(3) = 18, f'(3)=0 and f''(3) = 4. Then lim_(x → 1) logₑ [ f(2+x)/f(3)…

If lim_(x→ 0) (e^(a-1)x+2cos bx+(c-2)e^(-x))/(xcos x-logₑ(1+x)) =2, then a²+b²+c² is equal to

If the function f(x)=(e^x(e^(tan x - x)-1)+logₑ(sec x+tan x)-x)/(tan x-x) is continuous at x=0, then the value of f(0) is equal to

The value of lim_(x→ 0)(logₑ (sec(ex)· sec(e²x)… sec(e¹⁰x)))/(e²-e^(2cos x)) is equal to:

If lim_(x → ∞) ( e/(1 - e) ( 1/e - x/(1 + x)))^x = α, then the value of (logₑ α)/(1 + logₑ α) equals:

Evaluate the following limit: lim_(x → ∞) (2x² - 3x + 5) ( 3x - 1)^(x/2)/((3x² + 5x + 4) √(3x + 2)^x). The value of the limit is:

Evaluate the limit: lim_(x → 0) cscx ( √(2 cos²x + 3 cosx) - √(cos²x + sinx + 4)) is equal to:

Let f(x) = lim_(n → ∞) Σ_(r=0)ⁿ ( (tan ( x/2^(r+1)) + tan³ ( x/2^(r+1)))/(1 - tan² ( x/2^(r+1)))) Then, lim_(x → 0) (e^x - e^f(x))/(x -…

Let [t] be the greatest integer less than or equal to t. Then the least value of p ∈ ℕ for which lim_(x → 0⁺) ( x ( [ 1/x ] + [ 2/x ] + … +…

If lim_(t → 0) ( ∫₀¹ ( 3x + 5)^t dx)^(1/t) = α/5e ( 8/5)^(2/3), then α is equal to:

For α, β, γ ∈ ℝ, if lim_(x → 0) (x² sin α x + (γ - 1)e^x²)/(sin 2x - β x) = 3, then β + γ - α is equal to:

If lim_(x → 0) (cos(2x) + a cos(4x) - b)/x⁴ is finite, then (a + b) is equal to:

If lim_(x → 0) ( (tan x)/x)^(1/x²) = p, then 96 logₑ p is equal to

If lim_(x → 1⁺) (x-1)(6+λ cos(x-1)) + μ sin(1-x)/(x-1)³ = -1, where λ, μ ∈ ℝ, then λ + μ is equal to

Evaluate the following limit: lim_(x → 0⁺) (tan(5x^(1/3)) log(1 + 3x²))/(tan⁻¹(3√x))² (e^5x^(4/3) - 1)

If the function f(x) = (tan(tan x) - sin(sin x))/(tan x - sin x) is continuous at x = 0, then f(0) is equal to:

For t > -1, let α_t and β_t be the roots of the equation ( (t + 2)^(1/7) - 1)x² + ( (t + 2)^(1/6) - 1)x + ( (t + 2)^(1/21) - 1) = 0. If…

Given below are two statements: Statement I: lim_(x → 0) ( (tan⁻¹ x + logₑ √(1+x)/(1-x) - 2x)/x⁵) = 2/5 Statement II: lim_(x → 1) (…

If a = lim_(x→ 0)[(√(1 + √(1 + x⁴)) - √2)/x⁴] and b = lim_(x→ 0)[sin²(x)/(√2 - √(1 + cos x))], then ab³ is:

If lim_(x→ 0) (3 + α sin(x) + β cos(x) + log(1 - x))/3tan²(x) = 1/3, then 2α - β is equal to:

Evaluate the limit lim_(x → π/2) 1/(x - π/2)² ∫_x³^(π/2)³ cos(t^(1/3)) dt, which is equal to:

Let the slope of the line 45x + 5y + 3 = 0 be 27r₁ + 9r₂/2 for some r₁, r₂ ∈ ℝ. Then lim_(x → 3)((∫₃^x 8t² dt)/(3r₂x/2 - r₂x² - r₁x³ - 3x))…

Let f: (-π/2, π/2) → ℝ be a differentiable function such that f(0) = 1/2. If lim_(x → 0) (x ∫₀^x f(t) dt)/(e^x² - 1) = α, then 8α² is equal…

Evaluate the limit: lim_(x → 0) [(e^2|sin(x)| - 2|sin(x)| - 1)/x²]

If lim_(x → ∞) (f(7x)/f(x)) = 1, then lim_(x → ∞) [(f(5x)/f(x)) - 1] is:

If the function f(x) = (72^x - 9^x - 8^x + 1)/(√2 - √(1 + cos x)), x ≠ 0 [2ex] a logₑ 2 logₑ 3, x = 0 is continuous at x = 0, then the…

Let f(x) = ∫₀^x (t + sin(1 - e^t)) dt, x ∈ ℝ. Find lim_(x → 0) f(x)/x³:

If lim_n→∞ (1²-1)(n-1)+(2²-2)(n-2)+…+((n-1)²-(n-1))/(1³+2³+…+n³)-(1²+2²+…+n²) is equal to:

The value of lim_(n → ∞) (1 + 2 - 3 + 4 + 5 - 6 + … + (3n - 2) + (3n - 1) - 3n)/(√(2n⁴ + 4n + 3) - √(n⁴ + 5n + 4)) is:

Let x = 2 be a root of the equation x² + px + q = 0 and f(x) = (1 - cos(x² - 4px + q² + 8q + 16))/(x - 2p)⁴, x ≠ 2p, 0, x = 2p. Then lim_(x…

lim_(x → 0) 48/x⁴ ∫₀^x t³/(t⁶ + 1) dt is equal to.

lim_(n → ∞) ( 3n [ 4 + ( 2 + 1/n)² + ( 2 + 2/n)² + … + ( 3 - 1/n)² ])

The number of real roots of the equation √(x² - 4x + 3) + √(x² - 9) = √(4x² - 14x + 6) is:

Evaluate the limit: lim_(x → ∞) (( √(3x+1) + √(3x-1))⁶ + ( √(3x+1) - √(3x-1))⁶)/(( x + √(x²-1))⁶ + ( x - √(x²-1))⁶) · x³

If lim_(n → ∞) ( 1/(1+n) + 1/(2+n) + 1/(3+n) + … + 1/2n) is equal to:

lim_(n → ∞){(2^(1/2)-2^(1/3))(2^(1/2)-2^(1/5))…(2^(1/2)-2^(1/(2n+1)))} is equal to:

lim_(x → 0) ( (1 - cos²(3x))/cos³(4x))( sin³(4x)/(logₑ(2x + 1))⁵) is equal to:

Among (S1) lim_(n → ∞) 1/n² ( 2 + 4 + 6 + … + 2n) = 1 (S2) lim_(n → ∞) 1/n¹⁶ ( 1¹⁵ + 2¹⁵ + 3¹⁵ + … + n¹⁵) = 1/16

If lim_(x → 0) (e^ax - cos(bx) - (cx e^(-cx))/2)/(1 - cos(2x)) = 17, then 5a² + b² is equal to: