Class 11 Maths Chapter 12 – Limits and Derivatives (NCERT Solutions with Formulas & Examples)

Introduction

This chapter introduces Calculus, the study of change. We begin with the intuitive idea of derivatives, then define limits, explore algebra of limits, and finally move to derivatives of standard functions. These concepts form the foundation for higher mathematics and applications in physics, engineering, and economics.

Key Formulas

• Average velocity:

$$\text{v}=\frac{s(t_2)-s(t_1)}{t_2-t_1}$$

• Limit definition:

$$\underset{x\to a}{\lim }f(x)=L$$

• Important limits:

$$\underset{x\to 0}{\lim }\frac{\sin x}{x}=1,\underset{x\to 0}{\lim }\frac{1-\cos x}{x}=0$$

• Polynomial limit:

$$\underset{x\to a}{\lim }f(x)=f(a)$$

• Derivative definition:

$$f^{\mathrm{\prime }}(x)=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}$$

Solved NCERT Examples (Step by Step)

Example 1

Find ${\lim }_{x\to 1}(x^3-x^2+1)$. Solution: Substitute directly: $1^3-1^2+1=1$.

Example 2

Evaluate ${\lim }_{x\to 3}x(x+1)$. Solution: Substitute: $3(3+1)=12$.

Example 3

Evaluate ${\lim }_{x\to 1}\frac{x^{15}-1}{x-1}$. Solution: Use theorem: ${\lim }_{x\to a}\frac{x^n-a^n}{x-a}=n{a}^{n-1}$. Here, $n=15,a=1$. So limit = $15\cdot 1^{14}=15$.

Example 4

Evaluate ${\lim }_{x\to 0}\frac{\sqrt{1+x}-1}{x}$. Solution: Put $y=1+x$. Then limit = ${\lim }_{y\to 1}\frac{\sqrt{y}-1}{y-1}$. Apply theorem: derivative of $\sqrt{y}$ at y=1 → $\frac{1}{2}$.

Example 5

Evaluate ${\lim }_{x\to 0}\frac{\sin x}{x}$. Solution: Using geometric proof, limit = 1.

(Continue similarly for all NCERT solved examples from Exercises 12.1, 12.2, and Miscellaneous.)

15 FAQs with Step‑by‑Step Solutions 

Q1. What is the limit of f(x) = x+10 as x → 5? Answer: Substitute: f(5) = 15. So limit = 15.

Q2. Find ${\lim }_{x\to 1}{x}^3$. Answer: Substitute: $1^3=1$. Limit = 1.

Q3. Find ${\lim }_{x\to 2}3x$. Answer: Substitute: 3(2) = 6. Limit = 6.

Q4. Find ${\lim }_{x\to 2}3$. Answer: Constant function → limit = 3.

Q5. Find ${\lim }_{x\to 1}(x^2+x)$. Answer: Substitute: $1^2+1=2$. Limit = 2.

Q6. Find ${\lim }_{x\to \pi \mathrm{/}2}\sin x$. Answer: Substitute: sin(π/2) = 1. Limit = 1.

Q7. Find ${\lim }_{x\to 0}(x+\cos x)$. Answer: Substitute: 0 + cos(0) = 1. Limit = 1.

Q8. Find ${\lim }_{x\to 0^{+}}\frac{1}{x}$. Answer: As x → 0+, value → ∞. Limit = ∞.

Q9. Find ${\lim }_{x\to 0^{-}}(x-2)$. Answer: As x → 0 from left, value → -2. Limit = -2.

Q10. Find ${\lim }_{x\to 0^{+}}(x+2)$. Answer: As x → 0 from right, value → 2. Limit = 2.

Q11. Find ${\lim }_{x\to 1}\frac{x^2+1}{x+100}$. Answer: Substitute: (1+1)/(1+100) = 2/101.

Q12. Find ${\lim }_{x\to 2}\frac{x^3-4x^2+4x}{x^2-4}$. Answer: Simplify: numerator = x(x-2)², denominator = (x-2)(x+2). Cancel (x-2). At x=2 → 0.

Q13. Find ${\lim }_{x\to 2}\frac{x^3-2x^2}{x^2-5x+6}$. Answer: Simplify denominator = (x-2)(x-3). Cancel (x-2). At x=2 → 4/(-1) = -4.

Q14. Find ${\lim }_{x\to 0}\frac{\sin x}{x}$. Answer: Standard limit = 1.

Q15. Find ${\lim }_{x\to 0}\frac{1-\cos x}{x}$. Answer: Standard limit = 0.

For complete NCERT Class 11 Maths solutions, visit www.fuzymathacademy.com.