Limits and Derivatives
Board: CBSE | Class: Class 11
Comprehensive study notes for Limits and Derivatives by Ajay Yadav (Math King of Katargam).
Key Concepts
Limits
limx→a f(x) = L. Left-hand limit: x→a⁻. Right-hand limit: x→a⁺. Limit exists iff LHL = RHL.
Standard Limits
limx→0 sinx/x = 1. limx→0 (ex-1)/x = 1. limx→0 (ax-1)/x = lna. limx→0 (1+x)¹/x = e.
Limit Laws
lim(f+g) = lim f + lim g. lim(fg) = (lim f)(lim g). lim(f/g) = (lim f)/(lim g) if lim g ≠ 0.
Derivative from First Principles
f'(x) = limh→0 [f(x+h)-f(x)]/h. This is the definition of the derivative as the slope of the tangent.
Basic Derivatives
d/dx(xⁿ) = nxⁿ⁻¹. d/dx(sinx) = cosx. d/dx(cosx) = -sinx. d/dx(tanx) = sec²x. d/dx(ex) = ex. d/dx(lnx) = 1/x.
Algebra of Derivatives
(f+g)’ = f’+g’. (fg)’ = f’g+fg’ (product rule). (f/g)’ = (f’g-fg’)/g² (quotient rule). Chain rule: d/dx f(g(x)) = f'(g(x))g'(x).
Important Formulas
| Derivative definition | f'(x) = limh→0[f(x+h)-f(x)]/h |
| Power rule | d/dx(xⁿ) = nxⁿ⁻¹ |
| Product rule | (fg)' = f'g + fg' |
| Quotient rule | (f/g)' = (f'g-fg')/g² |
| Chain rule | dy/dx = (dy/du)(du/dx) |
Solved Examples
Example 1: Find limx→2 (x²-4)/(x-2).
Solution: = limx→2 (x-2)(x+2)/(x-2) = limx→2 (x+2) = 4.
Example 2: Find derivative of x³ from first principles.
Solution: limh→0((x+h)³-x³)/h = limh→0(3x²h+3xh²+h³)/h = 3x².
Example 3: Find d/dx(sin2x).
Solution: Using chain rule: cos(2x) × 2 = 2cos2x.
Practice Questions
- Find limx→0 sin5x/x.
- Find derivative of 1/x from first principles.
- Find d/dx(tanx secx).
- Find limx→0 (1-cosx)/x².
- If y = sin(x²), find dy/dx.
Download PDF
Click here to download PDF notes.
Video Lessons
Watch video explanations on our Videos page.