Question 1

Evaluate the limit as x approaches 0: ((1+tan x)/(1+sin x))^(csc x).

Accepted Answer

1. ln of base = ln(1+tanx)-ln(1+sinx) ~ (tanx - sinx) ~ x^3/2. Multiplying by cosec x ~ 1/x gives ~ x^2/2 -> 0, so the limit is e^0 = 1.

Question 2

Evaluate the limit as x approaches 0 of sin([x−3]) divided by [x−3], where [.] denotes the greatest integer function.

Accepted Answer

the limit does not exist. Near x=0, x-3 is near -3: for x->0- we get [x-3] = -4 (limit sin4/4) and for x->0+ we get [x-3] = -3 (limit sin3/3). The one-sided limits differ, so the limit does not exist.

Question 3

Evaluate the limit as x approaches 0: (sin(x⁴) − x⁴ cos(x⁴) + x²⁰) / (x⁴(e^(2x⁴) − 1 − 2x⁴)).

Accepted Answer

1/6. Let u = x^4. Numerator sin u - u cos u + ... ~ u^3/3 = x^12/3. Denominator x^4(e^(2u)-1-2u) ~ x^4 * 2u^2 = 2 x^12. The limit = (1/3)/2 = 1/6.

Question 4

Evaluate the limit lim_(n→∞)(n³-8)/(n³+8). What is its value?

Accepted Answer

None of these. As n approaches infinity, the dominant terms in the numerator and denominator are both n³. Thus, the limit simplifies to the ratio of the leading coefficients, which is 1. Therefore, the correct answer is 'None of these' since 1 is not listed as an option.

Question 5

If lim_(x→∞) (1+(λ)/(x)+(μ)/(x²))^(2x)=e², which of the following values of λ and μ is correct?

Accepted Answer

λ=1,μ can be any real number. As x->infinity, (1+lambda/x+mu/x^2)^(2x) -> exp(2 lambda). Setting e^(2 lambda)=e^2 gives lambda=1, and mu can be any real number.

Question 6

Consider the following statements:

Statement 1: If lim_(x→α)(sin(f(x)))/(x-α), where f(x)=ax²+bx+c, exists as a finite nonzero number, then lim_(x→α)(1/f(x)-1)/(1/f(x)+1) does not exist.

Statement 2: The limit lim_(x→α)(sin(f(x)))/(x-α) can be finite only in the indeterminate form 0/0.

Accepted Answer

Statement 1 is incorrect, Statement 2 is correct. For sin(f(x))/(x-alpha) to be finite, f(alpha)=0, so f->0 and 1/f->infinity. Then (1/f-1)/(1/f+1)->1, which exists, making Statement 1 false. Statement 2 is true since a finite limit requires the 0/0 form. Hence Statement 1 incorrect, Statement 2 correct.

Question 7

Evaluate the limit lim_(x→∞) ((3x-4)/(3x+2))^((x+1)/(3)):

Accepted Answer

e^(-2/3). Write (3x-4)/(3x+2)=1-6/(3x+2). The limit is exp(lim ((x+1)/3)*(-6/(3x+2))) = exp(-6/(3*3)) = exp(-2/3).

Question 8

Evaluate the limit lim_(x→ 0)(xtan(2x)-2xtan x)/((1-cos 2x)²).

Accepted Answer

1/2. As x->0, x tan2x - 2x tan x = x[(2x+8x^3/3)-2(x+x^3/3)] = x(2x^3)=2x^4 to leading order, and (1-cos2x)^2=(2x^2)^2=4x^4. Limit = 2/4 = 1/2.

Question 9

Evaluate the limit lim_(x→ 2)(√(1+√(2+x)-√(3)))/(x-2).

Accepted Answer

(1)/(8√(3)). Let g(x)=sqrt(1+sqrt(2+x)). Then g(2)=sqrt(1+2)=sqrt3 and g'(x) = 1/(2g) * 1/(2 sqrt(2+x)). At x=2: g' = 1/(2 sqrt3) * 1/(2*2) = 1/(8 sqrt3). So the limit = 1/(8 sqrt3) = sqrt(3)/24.

Question 10

Evaluate the limit lim_(x→ 0)((4^x-1)³)/(sin ((x²)/(4)) log(1+3x)).

Accepted Answer

(4)/(3)(ln 4)³. The limit evaluates to rac{4}{3}( ext{ln } 4)³ because the numerator approaches 0 at a cubic rate, while the denominator approaches 0 at a quadratic rate, leading to a limit that reflects the higher order of growth in the numerator.

Question 11

Evaluate the limit lim_(x→ 0)(1-cos³ x)/(xsin xcos x).

Accepted Answer

(3)/(2). Near 0, 1-cos x ~ x^2/2 so 1-cos^3 x ~ 3(x^2/2), while x sin x cos x ~ x^2. The limit is (3/2).

Question 12

Evaluate the limit as x approaches 0 of (√(1-cos(x²)))/(1-cos x).

Accepted Answer

√(2). The limit evaluates to sqrt{2} because as x approaches 0, both the numerator and denominator approach 0, allowing us to apply L'Hôpital's rule or use trigonometric identities to simplify the expression, ultimately leading to the result of sqrt{2}.

Question 13

Evaluate the limit as x approaches 0 of (cosec x) raised to the power 1/log x:

Accepted Answer

1/e. ln(value)=ln(cosec x)/ln x = -ln(sin x)/ln x. As x->0+, ln(sin x) ~ ln x, so the ratio -> -1 and the limit is e^(-1)=1/e.

Question 14

Evaluate the limit as x approaches 0 of (8/x⁸) times [1 - cos(x²/2) - cos(x²/4) + cos(x²/2)cos(x²/4)].

Accepted Answer

1/32. The bracket = (1-cos(x^2/2))(1-cos(x^2/4)). Using 1-cos(u)~u^2/2: 1-cos(x^2/2)~x^4/8 and 1-cos(x^2/4)~x^4/32, so the product ~x^8/256. Multiplying by 8/x^8 gives 8/256 = 1/32.

Question 15

Find the constants a and b such that
lim_(x→∞) ((x²+1)/(x+1)-ax-b)=0.
Which pair of values satisfies this condition?

Accepted Answer

a = 1, b = -1. (x^2+1)/(x+1) = x - 1 + 2/(x+1). For (expression - ax - b) -> 0 as x -> infinity, the leading and constant terms must cancel: a = 1, b = -1 (the remainder 2/(x+1) -> 0).

Question 16

For the function f(x) defined by f(x) = (cos x)^(1/x) for x ≠ 0 and f(0) = k, if f is continuous at x = 0, what must k be?

Accepted Answer

1. For the function to be continuous at x = 0, the limit of f(x) as x approaches 0 must equal f(0). Evaluating the limit, we find that it approaches 1, so k must be 1 to ensure continuity.

Question 17

If f(x) = ∑(n=0 to ∞) xⁿ/n! (log a)ⁿ, then at x = 0, f(x)

Accepted Answer

is differentiable. Sum x^n (log a)^n / n! = e^{x log a} = a^x, which is differentiable for all x (in particular at x=0). So f is differentiable.

Question 18

Consider the function f(x)=g(x) (e^(1/x)-e^(-1/x))/(e^(1/x)+e^(-1/x)), where g is continuous. The limit lim_(x→ 0) f(x) fails to exist when

Accepted Answer

g(x) is a constant function. The factor (e^(1/x)-e^(-1/x))/(e^(1/x)+e^(-1/x)) = tanh(1/x) tends to +1 as x->0+ and -1 as x->0-. So the limit of f exists only if g(0)=0. If g is a (non-zero) constant, the left and right limits are -g(0) and +g(0), which differ, so the limit fails to exist. For g(x)=x, x^2, or x*h(x) we have g(0)=0 and the limit exists.

Question 19

Let f(x) = tan(⌊x⌋π) / ⌊1 + |log(sin² x + 1)|⌋, where ⌊·⌋ denotes the greatest integer function and | | denotes the modulus, then f(x) is

Accepted Answer

continuous for every x. The numerator is tan(floor(x)*pi); since floor(x) is an integer n, tan(n*pi)=0 for all x. The denominator floor(1+|log(sin^2 x+1)|) stays equal to 1 because sin^2 x+1 in [1,2] gives log in [0,log2] and 1+that in [1,1.30]. Hence f(x)=0/1=0 for every x, which is continuous everywhere.

Question 20

For which values of a does the function f(x)=(a²-3a+2) ((cos² x)/(4)-(sin² x)/(4))+(a-1)x+sin 1 have no critical points?

Accepted Answer

(0,1)∪(1,4). f(x) = (a^2-3a+2)(cos2x)/4 + (a-1)x + sin1, so f'(x) = (a-1)[1 - (a-2)(sin2x)/2]. For no critical point f' must never be 0. Excluding a=1 (where f'==0), need |a-1| > |(a-1)(a-2)/2|, i.e. |a-2| < 2 -> 0

Question 21

Find the equation of the tangent to the graph of y = e^(- |x|) at the point where it intersects the vertical line x = 1.

Accepted Answer

None of these. The tangent line at the point where x = 1 must be calculated using the derivative of the function at that point. The correct equation does not match any of the provided options, confirming that 'None of these' is the right choice.

Question 22

For the function f(x) = (a² - 1)/(a² + 1)x³ - 3x + 5 to be decreasing for every real x, the admissible values of the parameter a are

Accepted Answer

[-1, 1]. The function is decreasing when its derivative is less than or equal to zero for all x. By analyzing the derivative, we find that the condition for it to be non-positive is satisfied when a is within the range [-1, 1], ensuring that the cubic term does not dominate and cause the function to increase.

Question 23

Evaluate the limit as n tends to infinity of the sum 1/(2n+1) + 1/(2n+2) +... + 1/(3n).

Accepted Answer

logₑ(3/2). Write the sum as (1/n) * sum over r=1..n of 1/(2+r/n), a Riemann sum for the integral from 0 to 1 of dx/(2+x) = ln(3/2). So the limit is ln(3/2).

Question 24

If lim_(x→ ∞) (1+(a)/(x)+(b)/(x²))^(2x)=e², then which of the following describes a and b?

Accepted Answer

a=1, with b any real number. The limit expression approaches the form of the exponential function, and for it to equal e², the coefficient of the linear term in the expansion must be 1, which means a must be 1. The term involving b does not affect the leading behavior as x approaches infinity, allowing b to be any real number.

Question 25

If α and β are the two different zeros of the quadratic ax² + bx + c = 0, then the value of

lim_(x→α) (1 - cos(ax² + bx + c))/((x - α)²)
∋s

Accepted Answer

a²/2 (α - β)². The limit evaluates the behavior of the function as x approaches one of its roots, α. By applying L'Hôpital's rule and recognizing that the expression inside the cosine approaches zero, we find that the limit simplifies to a form involving the derivative of the cosine function, leading to the result of a²/2 (α - β)².

Question 26

Let f: R → R be a positive, increasing function such that lim_(x→∞) (f(3x))/(f(x)) = 1. Then what is lim_(x→∞) (f(2x))/(f(x))?

Accepted Answer

1. The limit condition indicates that as x approaches infinity, the growth rate of f(3x) and f(x) becomes similar, suggesting that f(x) grows at a rate that is asymptotically linear. Therefore, by the same reasoning, the limit of f(2x) compared to f(x) will also approach 1, indicating that f(2x) and f(x) grow at the same rate as x approaches infinity.

Question 27

Let f: R → [0,∞) be a function for which the limit lim_(x→5) f(x) exists, and also lim_(x→5) ((f(x))² - 9)/(√(|x-5|)) = 0. Then the value of lim_(x→5) f(x) is:

Accepted Answer

3. The limit condition indicates that as x approaches 5, the expression (f(x))² approaches 9, which implies that f(x) approaches either 3 or -3. However, since f(x) is constrained to be non-negative (f: R → [0,∞)), the only feasible limit is 3.

Question 28

Evaluate the limit as x approaches 0: ((1-cos 2x)(3+cos x))/(xtan 4x).

Accepted Answer

2. As x approaches 0, the expression can be simplified using Taylor series expansions for cosine and tangent. The limit evaluates to 2 after applying L'Hôpital's rule or recognizing that the leading terms in the numerator and denominator yield a limit of 2.

Question 29

Evaluate the limit as x approaches 0: (sin(π cos² x))/(x²).

Accepted Answer

π. As x approaches 0, Bcos² x approaches 1, making Bsin(Bpi Bcos² x) approach Bsin(Bpi). Using L'Hôpital's rule or the small angle approximation, the limit simplifies to Bpi, confirming that the correct answer is Bpi.

Question 30

Evaluate the limit as x approaches π/2: (cot x - cos x) / (π - 2x)³.

Accepted Answer

1/16. The limit evaluates to 1/16 because as x approaches π/2, both the numerator and denominator approach 0, allowing us to apply L'Hôpital's Rule. After differentiating the numerator and denominator appropriately, the limit simplifies to 1/16.

Question 31

For every real number t, let [t] denote the greatest integer less than or equal to t. Then evaluate lim_(x→ 0⁺) x ([1/x]+[2/x]+⋯+[15/x]).

Accepted Answer

It is 120.. As x approaches 0 from the positive side, each term [k/x] for k = 1 to 15 approaches k/x, leading to the sum [1/x] + [2/x] +... + [15/x] approximating (1 + 2 +... + 15)(1/x) = 120/x. Multiplying by x gives the limit as 120, confirming that the correct answer is 120.

Question 32

Evaluate the limit as y approaches 0: (√(1+√(1+y⁴))-√(2))/y⁴.

Accepted Answer

The limit exists and is 1/(4√(2)). The limit can be evaluated using L'Hôpital's rule, which applies because both the numerator and denominator approach 0 as y approaches 0. After applying the rule and simplifying, the result converges to 1/(4√2), confirming that the limit exists.

Question 33

For y = (x + √(1 + x²))ⁿ, the value of (1 + x²) d²y/dx² + x dy/dx is

Accepted Answer

n²y. The expression (1 + x²)d²y/dx² + x dy/dx simplifies to n²y due to the properties of the function y and its derivatives, which reflect the structure of the original equation when differentiated appropriately.

Question 34

If x = e^(y + e^(y +...)), x > 0, then dy/dx is

Accepted Answer

(1 - x)/x. The inner expression e^(y+...) equals x itself, so x=e^(y+x). Taking logs: ln x = y + x, hence y = ln x - x and dy/dx = 1/x - 1 = (1 - x)/x.

Question 35

The set of points where f(x) = x/(1 + |x|) is differentiable is

Accepted Answer

(-∞, ∞). For x>0, f=x/(1+x) with f'=1/(1+x)^2; for x<0, f=x/(1-x) with f'=1/(1-x)^2. Both one-sided derivatives at 0 equal 1 and f is continuous, so f is differentiable everywhere -> domain (-inf, inf).

Question 36

The function f: R - {0} → R given by f(x) = 1/x - 2/(e^(2x) - 1) can be made continuous at x = 0 by defining f(0) as

Accepted Answer

1. Combine to [(e^(2x)-1)-2x]/[x(e^(2x)-1)]. With e^(2x)-1 = 2x+2x^2+(4/3)x^3+..., numerator = 2x^2+(4/3)x^3, denominator = 2x^2+2x^3, so the limit as x->0 is 1. Define f(0)=1.

Question 37

Let f be a function on the interval (π/6, π/3) given by
f(x) = { (√2 cos x − 1)/(cot x − 1), for x ≠ π/4; k, for x = π/4 }
If f is continuous, what is the value of k?

Accepted Answer

1/2. Write f(x)=sin x*(sqrt2 cos x -1)/(cos x - sin x). With x=pi/4+h, sqrt2 cos x -1 ~ -h, cos x - sin x ~ -sqrt2 h, sin x ~ 1/sqrt2, so the limit is (1/sqrt2)(-h)/(-sqrt2 h)=1/2. Hence k=1/2.

Question 38

If the equation aₙ xⁿ + aₙ₋₁xⁿ⁻¹ +.......... + a₁ x = 0, a₁ ≠ 0, n ≥ 2, has a positive root x = α, then the equation n aₙ xⁿ⁻¹ + (n-1)aₙ₋₁xⁿ⁻² +.......... + a₁ = 0 has a positive root, which is

Accepted Answer

smaller than α. Let P(x)=a_n x^n+...+a_1 x; then P(0)=0 and P(alpha)=0. The second polynomial is P'(x). By Rolle's theorem P' has a root in (0,alpha), i.e. a positive root smaller than alpha.

Question 39

lim(x→0) ((1 − cos 2x)(3 + cos x)) / (x tan 4x) is equal to

Accepted Answer

2. As x approaches 0, the limit can be evaluated using L'Hôpital's Rule or by simplifying the expression. The numerator approaches 0 while the denominator also approaches 0, allowing us to differentiate both parts. After simplification, the limit evaluates to 2.

Question 40

lim x→0 (sin(π cos² x) / x²) is equal to

Accepted Answer

π. As x approaches 0, cos²(x) approaches 1, making sin(π cos²(x)) approach sin(π), which is 0. Applying L'Hôpital's rule to the limit of the form 0/0, we differentiate the numerator and denominator, ultimately leading to the limit equating to π.

Question 41

Let p = lim x→0+ (1 + tan²(√x))^(1/2x) then log p is equal to

Accepted Answer

1/2. To find log p, we first analyze the limit as x approaches 0 from the right. The expression inside the limit simplifies to e^(lim x→0+ (1/2x) log(1 + tan²(√x))). As x approaches 0, tan²(√x) approaches 0, and using the Taylor expansion, we find that log(1 + tan²(√x)) behaves like tan²(√x). This leads to the limit evaluating to 1/2, thus log p equals 1/2.

Question 42

lim_(n→∞) ( ((n+1)(n+2)...3n)/(n²ⁿ))^(1/n) is equal to -

Accepted Answer

27/e². The limit evaluates the growth rate of the product in the numerator compared to the exponential growth in the denominator. By applying Stirling's approximation and simplifying the expression, we find that the dominant terms lead to the result of 27/e².

Question 43

lim x→0 ( (27 + x)^(1/3) − 3) / ( 9 − (27 + x)^(2/3)) equals -

Accepted Answer

−1/6. The limit evaluates to -1/6 because as x approaches 0, the expression simplifies using L'Hôpital's rule or algebraic manipulation, revealing that the rate of change in the numerator and denominator leads to this specific value.

Question 44

lim x→0 [x tan 2x − 2x tan x]/(1 − cos 2x)² equals -

Accepted Answer

1/2. (1-cos2x)^2 = 4 sin^4 x ~ 4x^4. Numerator x(tan2x - 2tanx) = x*2t^3/(1-t^2) ~ 2x^4 with t=tanx. So the limit = 2x^4 / 4x^4 = 1/2.

Question 45

Let f(x) = {(x−1)^(1/(2−x)), x > 1, x ≠ 2; k, x = 2}. The value of k for which f is continuous at x = 2 is -

Accepted Answer

e^−1. Let L = lim_{x->2} (x-1)^(1/(2-x)). Then ln L = lim (ln(x-1))/(2-x), a 0/0 form. By L'Hopital ln L = lim [1/(x-1)]/(-1) = -1. Hence k = L = e^-1.

Question 46

lim_(y→0) (√(1+√(1+y⁴))-√(2))/y⁴

Accepted Answer

exists and equals 1/(4√(2)). The limit can be evaluated using L'Hôpital's rule, which is applicable here since both the numerator and denominator approach zero as y approaches zero. After applying the rule and simplifying, the limit converges to 1/(4√2), confirming that the correct option is B.

Question 47

If lim x→1 (x⁴ − 1)/(x − 1) = lim x→k (x³ − k³)/(x² − k²), then k is:

Accepted Answer

8/3. The limit on the left can be simplified using L'Hôpital's rule or by factoring, resulting in 4. The limit on the right can also be simplified, and setting it equal to 4 allows us to solve for k, leading to the conclusion that k must be 8/3.

Question 48

lim x→π/4 (cot³ x − tan x) / cos(x + π/4) is:

Accepted Answer

8. As x approaches π/4, cot³(x) approaches 1 and tan(x) also approaches 1, making the numerator approach 0. The denominator, cos(x + π/4), approaches 0 as well, leading to an indeterminate form. Applying L'Hôpital's Rule, we differentiate the numerator and denominator, ultimately simplifying to find that the limit evaluates to 8.

Question 49

lim x→0 (sin²x)/(√2 − √(1 + cos x)) equals -

Accepted Answer

4√2. To evaluate the limit, we can use L'Hôpital's rule since both the numerator and denominator approach 0 as x approaches 0. After applying the rule and simplifying, we find that the limit equals 4√2.

Question 50

Let f: (−1, ∞) → R be defined by f(0) = 1 and f(x) = (1/x) logₑ (1 + x), x ≠ 0. Then the function f:

Accepted Answer

decreases in (−1, ∞). g(x)=ln(1+x)/x decreases throughout (-1, infinity): e.g. g(-0.9)=2.56, g(-0.5)=1.39, g(0)=1, g(1)=0.69. So f decreases in (-1, infinity).

JEE Main Maths: Limits and Derivatives questions with solutions

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