Limits — Chapter Test | Class 11 IMO

Q1

Find the value of lim(x→0) (tan 7x) / (2x).

Rewriting as (7/2) × (tan 7x) / (7x). Since lim(θ→0) (tan θ)/θ = 1, the limit is (7/2) × 1 = 7/2.

Q2

Calculate lim(x→0) [√(2 + x) − √2] / x.

Rationalise by multiplying numerator and denominator by [√(2 + x) + √2]. The limit simplifies to 1 / (√2 + √2) = 1/(2√2).

Q3

Find the left-hand limit of f(x) = |x − 3| / (x − 3) as x approaches 3.

For x < 3, |x − 3| = −(x − 3). So the expression is −(x − 3) / (x − 3) = −1. The left-hand limit is −1.

Q4

What is lim(x→0) [ 1 / x ] ?

As x approaches 0 from the right, 1/x approaches +∞. As x approaches 0 from the left, 1/x approaches −∞. Since LHL ≠ RHL, the limit does not exist.

Q5

The population of bacteria in a lab in Mumbai follows a curve where the instantaneous rate is given by lim(h→0) (2^(3+h) − 8) / h. What is this value?

Rewriting 2^(3+h) as 2³ × 2ʰ = 8 × 2ʰ. The expression is 8(2ʰ − 1)/h. As h approaches 0, (2ʰ − 1)/h approaches log(2). So the result is 8 log(2).

Q6

The value of lim(x→0) xsin(1/x) is:

Since |sin(1/x)| ≤ 1, |xsin(1/x)| ≤ |x|. By sandwich theorem, limit = 0.

Q7

Evaluate lim(x→0) (tan4x)/(tan2x).

Using tan(kx) ≈ kx near 0, limit = 4/2 = 2.

Q8

Determine the limit: lim(x→0) [ (1+x)⁶ − 1 ] / x.

Let y = 1 + x. As x→0, y→1. The limit becomes lim(y→1) (y⁶ − 1⁶) / (y − 1). Applying the standard formula, we get 6 × 1⁵ = 6.

Q9

Evaluate lim(x→0) (1 − cosx)/(sinx).

Numerator behaves like x²/2 and denominator like x. Ratio tends to 0.

Q10

Aarav is tracking the depreciation of a car. The value factor approaches lim(t→0) [log(1 - t)] / t. Find this limit.

Let u = -t. As t approaches 0, u approaches 0. The limit becomes lim(u→0) [log(1 + u)] / (-u) = -1 × lim(u→0) [log(1 + u)] / u = -1.

Q11

If f(x)=5 for all x≠2, then lim(x→2) f(x) equals:

A constant function has limit equal to the constant value.

Q12

Evaluate lim(x→0) (sin²x)/x².

[(sinx)/x]² → 1² = 1.

Q13

A bus fare function f(x)=2x+10 depends on distance x. The fare as distance approaches 20 km is:

Substitute x=20. Fare = 2×20+10 = ₹50.

Q14

If lim(x→1) [x⁴ − 1] / [x − 1] = lim(x→k) [x³ − k³] / [x² − k²], find the value of k.

LHS: 4(1)³ = 4. RHS: Divide numerator and denominator by (x−k) to get [3k²] / [2k] = (3/2)k. Equating them: 4 = (3/2)k, so k = 8/3.

Q15

Evaluate lim(x→2) 5.

The limit of a constant function is the constant itself. Therefore, lim(x→2) 5 = 5.