NCERT Plus One (Class 11) Mathematics • Chapter 12

Limits and Derivatives

Q: Evaluate lim_{x→1} (x¹⁵ − 1) / (x¹⁰ − 1).

Divide numerator and denominator by (x − 1) : lim_{x→1} (x¹⁵−1)/(x−1) (x¹⁰−1)/(x−1) = 15(1)¹⁴ 10(1)⁹ = 15 10 = 3 2 .

Q: Evaluate lim_{x→0} (sin ax + bx) / (ax + sin bx), where a, b, a + b ≠ 0.

Divide numerator and denominator by x : lim_{x→0} (sin ax)/x + b a + (sin bx)/x = a(1) + b a + b(1) = a + b a + b = 1 .

Q: Evaluate lim_{x→0} (√(1 + x) − 1) / x.

Method 1 (Rationalising): Multiply top & bottom by √(1 + x) + 1 : lim_{x→0} (1 + x) − 1 x(√(1 + x) + 1) = lim_{x→0} 1 √(1 + x) + 1 = 1 2 .

Q: Find lim_{x→0} f(x) where f(x) = 2x + 3 for x ≤ 0 and f(x) = 3(x + 1) for x > 0.

• LHL = lim_{x→0⁻} (2x + 3) = 2(0) + 3 = 3 . • RHL = lim_{x→0⁺} 3(x + 1) = 3(0 + 1) = 3 . Since LHL = RHL = 3 , lim_{x→0} f(x) = 3 .

Q: Find the derivative of f(x) = cos x from First Principles.

f'(x) = lim_{h→0} cos(x + h) − cos x h = lim_{h→0} −2 sin(x + h/2) sin(h/2) h = − lim_{h→0} sin(x + h/2) · sin(h/2) h/2 = −sin(x) · 1 = −sin x .

Q: Find the derivative of f(x) = 1 / x from First Principles.

f'(x) = lim_{h→0} 1/(x + h) − 1/x h = lim_{h→0} x − (x + h) h · x(x + h) = lim_{h→0} −h h · x(x + h) = lim_{h→0} −1 x(x + h) = − 1 x² .

Q: Differentiate y = (sin x + cos x) / (sin x − cos x) with respect to x.

Let u = sin x + cos x ⇒ u' = cos x − sin x ; v = sin x − cos x ⇒ v' = cos x + sin x . y' = (cos x − sin x)(sin x − cos x) − (sin x + cos x)(cos x + sin x) (sin x − cos x)² Numerator = −(sin x − cos x)² − (sin x + cos x)² = −2(sin²x + cos²x) = −2 . • Answer: −2 / (sin x − cos x)² .

Q: Find the derivative of (a) tan x using Quotient Rule and (b) x⁵ (3 − 6x⁻⁹) using Power Rule.

• (a) Write tan x = sin x / cos x : ( sin x cos x )' = cos²x − sin x(−sin x) cos²x = 1 cos²x = sec²x . • (b) Expand first: y = 3x⁵ − 6x⁻⁴ . y' = 15x⁴ − 6(−4)x⁻⁵ = 15x⁴ + 24x⁻⁵ .

Left/right-hand limits, algebraic & sin(x)/x squeeze limits, first-principles slope, power, product, quotient & trig derivatives.

📄 Official NCERT Textbook PDF

Topic 12.2: Intuitive Need for Limits & Left/Right-Hand Limits

Imagine cycling along a smooth Kochi bridge toward milepost x = 2, where a tiny manhole tile is missing (so standing AT x = 2 is undefined!). Still, cyclists approaching from the left (1.9, 1.99, 1.999) and from the right (2.1, 2.01, 2.001) BOTH zoom smoothly toward height 4! That expected target height — whether the tile exists or not — is a Limit!

WHAT IS A LIMIT?lim_{x→a} f(x) = L asks where f(x) heads as x gets super close to a (x ≠ a!).

LHL & RHLLHL: lim_{x→a⁻} f(x) (from left < a)
RHL: lim_{x→a⁺} f(x) (from right > a)

EXISTENCE RULELimit exists ONLY when:
LHL = RHL = finite L!

🧠 Memory Trick: A Limit asks "Where do the left road and right road aim to meet?" Even if there's an open hole at x = a, as long as Left meets Right, the limit EXISTS! If the road has a cliff jump (LHL ≠ RHL), the limit DOES NOT EXIST!

💡 Simple Kid-Check: To check lim_{x→a} f(x), cover the exact spot x = a with your thumb! Look only at what happens just to the left (a − h) and just to the right (a + h) as h → 0. If both fingers point to the SAME height L, that's your limit!

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Removable Hole Example: Let f(x) = x² − 4x − 2. At x = 2, it gives 0/0 (undefined!). Yet for x ≠ 2, f(x) = x + 2 → both LHL = 4 and RHL = 4, so lim_{x→2} f(x) = 4!
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Jump Discontinuity Example: For f(x) = |x|x at x = 0: for x < 0, f(x) = −1 (LHL = −1); for x > 0, f(x) = +1 (RHL = +1). Since LHL ≠ RHL, the limit at 0 does NOT exist!

Classic 1-mark & 2-mark DHSE trap: assuming lim_{x→a} f(x) = f(a) always! If a piecewise function sets f(x) = x + 2 for x ≠ 0 and f(0) = 99, then f(0) = 99, yet lim_{x→0} f(x) = 2! Always check both LHL and RHL around piecewise breakpoints (like x ≤ 0 vs x > 0)!

Topic 12.3: Algebra of Limits & Polynomial/Rational Evaluation (xⁿ−aⁿ)/(x−a)

When you plug x = 2 straight into (x² − 4)/(x − 2), you hit the dreaded 0/0 roadblock! Does 0/0 mean "no answer"? Never — 0/0 simply means a troublemaker factor (x − 2) is hiding upstairs and downstairs! Cancel it (or use our superpower Power Limit Formula) and the true limit pops right out!

ALGEBRA OF LIMITSlim(f ± g) = lim f ± lim g
lim(f·g) = lim f · lim g, lim(f/g) = lim f / lim g (lim g ≠ 0)

0/0 RESCUE KIT1) Factor & cancel (x − a)
2) Rationalize √ conjugate
3) Use xⁿ − aⁿ formula!

STAR THEOREM 2lim_{x→a} xⁿ − aⁿx − a = n · aⁿ⁻¹
(true for rational n too!)

🧠 Memory Trick: Look at lim_{x→a} (xⁿ − aⁿ)/(x − a) = n aⁿ⁻¹ — it matches the Power Rule of derivatives! Bring power n down in front and drop power by 1!

💡 Simple Kid-Check: For plain polynomials like 3x² − 5x + 1, just plug x = a directly! Only when plugging x = a gives 0/0 do you need to clean up the fraction by factoring, rationalizing, or using Theorem 2!

Super Useful Ratio Shortcut: lim_{x→a} xᵐ − aᵐxⁿ − aⁿ = mn · aᵐ⁻ⁿ (divide top & bottom by x − a!)

⚡ Two Quick Board-Style Examples:
1) lim_{x→1} x¹⁵ − 1x¹⁰ − 1 = divide top & bottom by (x − 1) → 15(1)¹⁴10(1)⁹ = 1510 = 32!
2) lim_{x→0} √(1 + x) − 1x → write t = 1 + x → 1: lim_{t→1} t¹/² − 1¹/²t − 1 = 12(1)⁻¹/² = 12!

Watch out for fake Theorem 2 traps like lim_{x→2} x³ + 8x − 2! Notice the numerator is PLUS 8, NOT minus 8 — plugging x = 2 gives 16/0, NOT 0/0! Only when MINUS appears (or xⁿ − (−a)ⁿ for odd n as x → −a!) does Theorem 2 apply! Always check for 0/0 FIRST!

Topic 12.4: Sandwich Intuition & Important Trigonometric Limits (sin x / x → 1)

Imagine making a tight Malabar egg sandwich: bottom slice of bread is cos x, top slice is 1, and tucked right in between is sin xx! As x → 0, the bottom slice cos 0 rises straight to 1, squeezing the middle slice cleanly to 1!

SANDWICH THEOREMIf f(x) ≤ g(x) ≤ h(x) & lim f = lim h = L, then lim g(x) = L!

TRIG LIMIT GEMSlim_{x→0} sin xx = 1
lim_{x→0} 1 − cos xx = 0

HANDY EXTENSIONSlim_{x→0} tan xx = 1
lim_{x→0} sin axbx = ab, 1−cos xx² → 12

🧠 Memory Trick: When angle x → 0 (in RADIANS!), sin x ≈ x and tan x ≈ x, so matching sine-over-same-angle always collapses to 1!

💡 Simple Kid-Check: Why does sin x / x → 1? On a tiny wedge of a unit circle (radius = 1), the straight height sin x and curved crust length x (in radians!) become practically identical as the slice gets razor-thin!

Key Inequality (0 < x < π/2): cos x < sin xx < 1 → as x → 0, both outer ends hit 1!

🥪 Fast Board-Exam Tricks:
• lim_{x→0} sin 4xsin 2x = lim_{x→0} (sin 4x / 4x) · 4x(sin 2x / 2x) · 2x = 1 · 41 · 2 = 2!
• lim_{x→0} 1 − cos xx = lim 2 sin²(x/2)x = lim sin(x/2)x/2 · sin(x/2) = 1 · 0 = 0!

Watch TWO sneaky Kerala HSE traps! (1) Degrees vs Radians: lim_{x→0} sin x°x is NOT 1! Convert x° = πx180 radians first → answer is π/180! (2) Check where x approaches: lim_{x→0} sin xx = 1, whereas if x → ∞, sin x / x → 0!

Topic 12.5: Derivative as Instantaneous Slope & First Principles Definition

Drive from Thrissur to Ernakulam: your average speed for the 2-hour trip might be 40 km/h, yet the instant you pass a speed camera at t = 1 hr, the speedometer flashes 72 km/h! Shrinking the measurement time window h → 0 turns average rate (secant slope) into exact instantaneous rate (tangent slope) — the Derivative!

FIRST PRINCIPLESf'(x) = lim_{h→0} f(x + h) − f(x)h
Also written dydx

AT A POINT af'(a) = lim_{h→0} f(a + h) − f(a)h
= Slope of tangent at (a, f(a))!

MEANINGSecant chord PQ rotates cleanly into the Tangent at P as h → 0!

🧠 Memory Trick: Derivative is literally Rise ÷ Run [f(x+h) − f(x)] / h in the limit where the tiny step h shrinks down to zero!

💡 Simple Kid-Check: Pick point P(a, f(a)) and a neighbour Q(a+h, f(a+h)) on a roller-coaster track. The straight bar connecting P and Q is a secant. Slide Q closer and closer toward P (h → 0) — the bar settles perfectly flush against the track at P as the tangent line!

🔬 Two Guaranteed First-Principles Derivations (3–4 Marks!):
1) f(x) = x²:
f'(x) = lim_{h→0} (x+h)² − x²h = lim_{h→0} 2xh + h²h = lim_{h→0} (2x + h) = 2x.

2) f(x) = sin x (Kerala HSE Favourite!):
f'(x) = lim_{h→0} sin(x+h) − sin xh = lim_{h→0} 2 cos(x + h/2) sin(h/2)h
Move 2 down to get sin(h/2)h/2 → 1 and cos(x + 0) = cos x → cos x!

When a Kerala DHSE question says "Find the derivative of ... from FIRST PRINCIPLES (ab initio)", using the Power Rule or direct formula directly earns ZERO marks! You MUST write f'(x) = lim_{h→0} [f(x+h) − f(x)] / h, cancel h cleanly before setting h = 0, and keep writing lim_{h→0} until the very step where h disappears!

Topic 12.5: Power Rule & Algebra of Derivatives (Product & Quotient Rules)

Doing First Principles every single time is like baking bread from wheat flour every breakfast! Once we prove the shortcuts, differentiating becomes lightning fast: drop the exponent with the Power Rule, juggle two multiplied functions with the Product Rule, and tame fractions with the Quotient Rule!

POWER & CONSTANTddx(xⁿ) = n·xⁿ⁻¹
ddx(c) = 0, ddx(x) = 1

PRODUCT RULE (uv)'(uv)' = u'v + uv'
(Take turns differentiating one!)

QUOTIENT RULE (u/v)'(uv)' = u'v − uv'v²
(u'v FIRST, minus, over v²!)

🧠 Memory Trick for Quotient Rule: "Low d-High minus High d-Low, square the bottom and away you go!" — i.e. (v·u' − u·v') / v²! Notice u'v is ALWAYS positive!

💡 Simple Kid-Check: Why isn't (u · v)' = u' · v'? Because when BOTH width u and height v of a rectangle grow slightly, the new area comes from TWO strips: u'·v + u·v'!

Handy Power-Rule Gems:
• ddx(√x) = ddx(x¹/²) = 12√x | ddx(1x) = ddx(x⁻¹) = −1x²

⚙️ Step-by-Step Quotient Rule Example: Differentiate y = x + 1x − 1.
Let u = x + 1 ⇒ u' = 1 and v = x − 1 ⇒ v' = 1.
dydx = (1)(x − 1) − (x + 1)(1)(x − 1)² = x − 1 − x − 1(x − 1)² = −2(x − 1)²!

Watch THREE common algebra traps: (1) Writing (u · v)' = u' · v' or (u/v)' = u'/v' — COMPLETELY FALSE! (2) Flipping the numerator order in Quotient Rule to (uv' − u'v)/v² (gives the wrong sign!). (3) Differentiating 1x² as 12x — NO! Rewrite as x⁻² first → −2x⁻³ = −2/x³!

Topic 12.5: Derivatives of Polynomial & Standard Trigonometric Functions

Watch a sine wave ripple across Kochi backwaters! Whenever the wave climbs fastest at x = 0, its slope hits +1 (cos 0); at its crest x = π/2, the surfboard levels flat with slope 0 (cos π/2); going downhill at x = π, slope hits −1 (cos π)! Every slope point of sin x literally draws cos x!

SIN & COS(sin x)' = cos x
(cos x)' = −sin x

TAN & COT(tan x)' = sec²x
(cot x)' = −cosec²x

SEC & COSEC(sec x)' = sec x tan x
(cosec x)' = −cosec x cot x

🧠 Memory Trick: Every function starting with "co-" (cos, cot, cosec) gets a MINUS (−) sign in its derivative! All others stay PLUS!

💡 Simple Kid-Check: You don't even have to memorize tan x separately! Write tan x = sin xcos x and run the Quotient Rule: (cos x)(cos x) − (sin x)(−sin x)cos²x = cos²x + sin²xcos²x = 1cos²x = sec²x!

Function `f(x)`	Derivative `f'(x)`	Starts with "co-"?	Quick Check at `x = 0`
`sin x`	`cos x`	No (+)	`1`
`cos x`	`−sin x`	Yes (−)	`0`
`tan x`	`sec²x`	No (+)	`1`
`cot x`	`−cosec²x`	Yes (−)	undefined
`sec x`	`sec x tan x`	No (+)	`0`
`cosec x`	`−cosec x cot x`	Yes (−)	undefined

🎯 Board-Style Mix: Differentiate y = x⁴ (5 sin x − 3 cos x) using Product Rule:
dydx = 4x³(5 sin x − 3 cos x) + x⁴(5 cos x − 3(−sin x)) = x³[20 sin x − 12 cos x + 5x cos x + 3x sin x]!

Watch out for TWO classic mistakes! (1) Putting a minus sign on sin x instead of cos x — remember the "co-" minus rule (cos, cot, cosec get minus; sin stays +cos!). (2) Differentiating cos(x) when x is in degrees, or differentiating sin²x as cos²x (FALSE! Write sin x · sin x and use Product Rule → 2 sin x cos x = sin 2x!).

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March • 1 Marks): Evaluate lim_{x→1} (x¹⁵ − 1) / (x¹⁰ − 1).

Divide numerator and denominator by (x − 1):
lim_{x→1} (x¹⁵−1)/(x−1)(x¹⁰−1)/(x−1) = 15(1)¹⁴10(1)⁹ = 1510 = 32.

Q2 (Kerala HSE Say • 2 Marks): Evaluate lim_{x→0} (sin ax + bx) / (ax + sin bx), where a, b, a + b ≠ 0.

Divide numerator and denominator by x:
lim_{x→0} (sin ax)/x + ba + (sin bx)/x = a(1) + ba + b(1) = a + ba + b = 1.

Q3 (Kerala HSE March • 2 Marks): Evaluate lim_{x→0} (√(1 + x) − 1) / x.

Method 1 (Rationalising): Multiply top & bottom by √(1 + x) + 1:
lim_{x→0} (1 + x) − 1x(√(1 + x) + 1) = lim_{x→0} 1√(1 + x) + 1 = 12.

Q4 (Kerala HSE March • 2 Marks): Find lim_{x→0} f(x) where f(x) = 2x + 3 for x ≤ 0 and f(x) = 3(x + 1) for x > 0.

• LHL = lim_{x→0⁻} (2x + 3) = 2(0) + 3 = 3.
• RHL = lim_{x→0⁺} 3(x + 1) = 3(0 + 1) = 3.
Since LHL = RHL = 3, lim_{x→0} f(x) = 3.

Q5 (Kerala HSE Say • 3 Marks): Find the derivative of f(x) = cos x from First Principles.

f'(x) = lim_{h→0} cos(x + h) − cos xh = lim_{h→0} −2 sin(x + h/2) sin(h/2)h
= − lim_{h→0} sin(x + h/2) · sin(h/2)h/2 = −sin(x) · 1 = −sin x.

Q6 (Kerala HSE March • 3 Marks): Find the derivative of f(x) = 1 / x from First Principles.

f'(x) = lim_{h→0} 1/(x + h) − 1/xh = lim_{h→0} x − (x + h)h · x(x + h)
= lim_{h→0} −hh · x(x + h) = lim_{h→0} −1x(x + h) = −1x².

Q7 (Kerala HSE Model • 3 Marks): Differentiate y = (sin x + cos x) / (sin x − cos x) with respect to x.

Let u = sin x + cos x ⇒ u' = cos x − sin x; v = sin x − cos x ⇒ v' = cos x + sin x.
y' = (cos x − sin x)(sin x − cos x) − (sin x + cos x)(cos x + sin x)(sin x − cos x)²
Numerator = −(sin x − cos x)² − (sin x + cos x)² = −2(sin²x + cos²x) = −2.
• Answer: −2 / (sin x − cos x)².

Q8 (NCERT / DHSE Model • 4 Marks): Find the derivative of (a) tan x using Quotient Rule and (b) x⁵ (3 − 6x⁻⁹) using Power Rule.

• (a) Write tan x = sin x / cos x:
(sin xcos x)' = cos²x − sin x(−sin x)cos²x = 1cos²x = sec²x.
• (b) Expand first: y = 3x⁵ − 6x⁻⁴.
y' = 15x⁴ − 6(−4)x⁻⁵ = 15x⁴ + 24x⁻⁵.