📘 Calculus · Chapter MN11🎯 NDA Level : High Priority
Limits, Continuity and Differentiability is the gateway to calculus — every subsequent topic in calculus builds on these three ideas. For NDA, the chapter is formula-rich and concept-driven. Questions test standard limit evaluations, checking continuity conditions at a point, and understanding the relationship between differentiability and continuity.
📌 What to expect in NDA (based on 2022–2024 papers): (1) Evaluating limits using factorisation (0/0 form), L’Hôpital’s rule, and standard results;
(2) Applying limx→0 (sin x)/x = 1 and related trigonometric limits;
(3) Checking the three conditions for continuity at a point;
(4) Finding k for a piecewise function to be continuous;
(5) Identifying removable, jump, or infinite discontinuities;
(6) Applying the first-principles definition of derivative;
(7) Differentiability vs continuity — which implies which;
(8) Points where |x|, x|x|, etc. are continuous but not differentiable.
Topics at a Glance
① Concept of Limit
limx→a f(x), left & right limits
② Limit Evaluation Methods
Factorisation, rationalisation, standard forms
③ Standard Limits
sin x/x, (1+1/n)n, etc.
④ Continuity
3 conditions, types of discontinuity
⑤ Differentiability
First principles, LHD, RHD
⑥ Continuity vs Differentiability
Implications, |x| example, corners
1. Concept of a Limit
1.1
What is a Limit? — Intuitive & Formal Definition
The limit is what f(x) APPROACHES, not necessarily what it equals at x = a
The limit of a function f(x) as x approaches a is the value that f(x) gets arbitrarily close to as x gets closer and closer to a — without x actually equalling a.
limx→a f(x) = L
Read: “the limit of f(x) as x approaches a equals L”
L is the value f(x) gets arbitrarily close to, but x ≠ a during this process
⚡ Left-Hand & Right-Hand Limits
Left-Hand Limit (LHL): lim[x→a⁻] f(x) = lim[h→0⁺] f(a - h)
x approaches a from the LEFT (values less than a)
Right-Hand Limit (RHL): lim[x→a⁺] f(x) = lim[h→0⁺] f(a + h)
x approaches a from the RIGHT (values greater than a)
Limit EXISTS if and only if:
LHL = RHL = L (both one-sided limits exist and are equal)
If LHL ≠ RHL: the limit does NOT exist (written DNE)
The limit of f(x) as x→a has NOTHING to do with f(a). The function may not even be defined at x = a, yet the limit can still exist.
Fig 1: Limit exists (LHL = RHL = L) even when f(a) is different. Open circle shows the limit value; filled circle shows the actual function value.
2. Methods of Evaluating Limits
2.1
Direct Substitution, Factorisation & Rationalisation
Try direct substitution first — if you get 0/0, use a simplification method
⚡ Step-by-Step Strategy for Any Limit
Step 1 — Direct substitution:
Put x = a directly. If you get a finite number → DONE.
If you get 0/0 or ∞/∞ → it is an INDETERMINATE FORM, continue.
Step 2 — Factorisation (for polynomial/rational 0/0 forms):
Factorise numerator and denominator separately.
Cancel the common factor (x - a).
Then substitute x = a.
Step 3 — Rationalisation (for square root 0/0 forms):
Multiply numerator and denominator by the conjugate.
Conjugate of (√x - √a) is (√x + √a).
Step 4 — Standard limit substitution:
Recognise the form and apply a known standard limit result.
Step 5 — L’Hôpital’s Rule (0/0 or ∞/∞):
lim f(x)/g(x) = lim f’(x)/g’(x) (differentiate top and bottom separately)
For NDA: Factorisation covers most algebraic limits. Standard limits cover trigonometric and exponential limits. L’Hôpital’s Rule is a fallback for any indeterminate form.
= limx→0 [(1+x)−1] / [x(√(1+x)+1)] = limx→0 x / [x(√(1+x)+1)]
= limx→0 1/(√(1+x)+1) = 1/(1+1) = 1/2.
2.2
Standard Limits — The Must-Memorise Results
These appear almost every year in NDA — know all forms cold
⚡ Standard Limit Results (All for NDA)
TRIGONOMETRIC LIMITS (x in radians):
lim[x→0] (sin x) / x = 1 ← most important
lim[x→0] (tan x) / x = 1
lim[x→0] (1 - cos x) / x = 0
lim[x→0] (1 - cos x) / x² = 1/2
lim[x→0] sin(ax) / sin(bx) = a/b (useful for related forms)
lim[x→0] sin(ax) / (bx) = a/b
EXPONENTIAL & LOGARITHMIC LIMITS:
lim[x→0] (eˣ - 1) / x = 1
lim[x→0] (aˣ - 1) / x = logₖ a (log to base e)
lim[x→0] log(1 + x) / x = 1
lim[x→∞] (1 + 1/x)ˣ = e
lim[x→0] (1 + x)^(1/x) = e
ALGEBRAIC LIMIT:
lim[x→a] (xⁿ - aⁿ) / (x - a) = n ⋅ a^(n-1) (for any rational n)
Memory trick for trig limits: Think of sin x ≈ x for very small x (in radians). So sin x / x ≈ x/x = 1 as x→0. All trig standard limits follow from this idea.
Always massage the form to match lim sin(u)/u where u→0
Algebraic Standard Limit
lim[x→1] (x⁵−1)/(x−1) = 5⋅1⁴ = 5
lim[x→2] (x⁵−32)/(x−2) = 5⋅2⁴ = 80
lim[x→a] (x^n−a^n)/(x−a) = n⋅a^(n-1)
Works for all rational n — including fractions and negatives
⚠ Critical: Degrees must match in trigonometric limits.
lim[x→0] sin(x²)/x = lim sin(x²)/x² ⋅ x → 1 ⋅ 0 = 0 (NOT 1).
lim[x→0] sin(x²)/x² = 1 (here argument x²→0 as x→0, so form is correct).
The standard result is limu→0 sin(u)/u = 1, where u must approach 0.
Answer: (c) 1/2
Standard result: limx→0(1−cosx)/x² = 1/2.
Derivation: use 1−cosx = 2sin²(x/2), then divide by x²: 2⋅[sin(x/2)/(x/2)]²⋅(1/4)⋅4 = 2⋅1⋅(1/4)⋅4 = ... = 1/2.
🔥 TRICKY QUESTIONS
Limits — Indeterminate Form Traps
🤯 T1. Evaluate limx→0 x / |x|.
LHL: limx→0− x/|x| = limx→0− x/(−x) = −1.
RHL: limx→0+ x/|x| = limx→0+ x/x = +1.
LHL ≠ RHL → limit does NOT exist. Trap: Students substitute 0 directly and get 0/0, then say the limit is 1. Must check one-sided limits for |x| problems.
🤯 T2. Evaluate limx→∞ (√(x²+x) − x).
Direct sub: ∞−∞ ← indeterminate. Rationalise:
= lim [(√(x²+x)−x)(√(x²+x)+x)] / (√(x²+x)+x)
= lim [(x²+x)−x²] / (√(x²+x)+x)
= lim x / (√(x²+x)+x). Divide by x:
= lim 1 / (√(1+1/x)+1) = 1/(1+1) = 1/2. Key strategy: for ∞−∞ with square roots, always rationalise first.
🤯 T3. Find limx→0 [sin(x²) + sin²(x)] / x².
= limx→0 sin(x²)/x² + limx→0 sin²(x)/x²
= limx→0 [sin(x²)/x²] + [limx→0 sin(x)/x]²
= 1 + 1² = 2. Note: sin(x²)/x² → 1 because the argument x²→0, matching the standard form sin(u)/u→1. Many students incorrectly split sin(x²) as (sin x)².
3. Continuity at a Point
3.1
Three Conditions for Continuity
ALL three must hold simultaneously — failing even one means discontinuous
A function f(x) is continuous at x = a if it satisfies all three conditions simultaneously. Think of it as: “you can draw the graph through x = a without lifting your pen.”
C1
f(a) is defined The function must have a value at x = a (no hole or undefined)
C2
limx→a f(x) exists LHL = RHL (both one-sided limits must be equal)
C3
limx→a f(x) = f(a) The limit value must equal the actual function value
⚡ Continuity Condition — Compact Form
f is continuous at x = a ⇔ lim[x→a] f(x) = f(a)
This single equation implicitly requires ALL THREE conditions:
— f(a) must exist (for RHS to be defined)
— lim must exist (LHL = RHL)
— They must be equal
For piecewise functions:
Check LHL = lim[x→a⁻] f(x) and RHL = lim[x→a⁺] f(x)
And verify both equal f(a)
For piecewise functions in NDA: find the value of the unknown constant k by setting LHL = f(a) = RHL. This is the most common NDA problem type in continuity.
Worked Example — Finding k for Continuity
f(x) = kx² for x ≤ 2, and f(x) = 3 for x > 2. Find k so that f is continuous at x = 2.
Classified based on what condition is violated and how
Removable Discontinuity
limx→a f(x) exists but does not equal f(a) (or f(a) is undefined). The gap can be “filled” by redefining f(a) = L.
Jump Discontinuity
LHL ≠ RHL (both finite, but different). The graph “jumps” from one value to another. Cannot be removed by redefining one point.
Infinite Discontinuity
At least one one-sided limit is ±∞. Common in 1/x type functions at x = 0. Also called essential discontinuity.
Visual — Types of Discontinuity on a Graph
Fig 2: Three types of discontinuity — Removable (hole), Jump (step), Infinite (vertical asymptote).
📌 Continuity of standard functions (memorise):
• Polynomials: continuous everywhere. • Rational f(x)/g(x): continuous where g(x) ≠ 0.
• sin x, cos x: continuous everywhere. • tan x: discontinuous at x = (2n+1)π/2.
• |x|: continuous everywhere (but not differentiable at x = 0).
• ex, log x (x>0): continuous on their domains.
📝 TOPIC-WISE PYQ
Continuity — NDA-Pattern Questions
Q5. Find k so that f(x) = (x²−9)/(x−3) for x≠3, and f(3)=k, is continuous at x=3.
(a) 3 (b) 6 (c) 9 (d) 0
Answer: (b) 6
limx→3(x²−9)/(x−3) = limx→3(x+3) = 6.
For continuity: f(3) = k = lim = 6.
Q6. f(x) = sin(x)/x for x ≠ 0 and f(0) = 2. Then f is:
(a) Continuous at x=0 (b) Discontinuous at x=0 (c) Not defined at x=0 (d) Continuous everywhere
Answer: (b) Discontinuous at x=0
limx→0 sin(x)/x = 1, but f(0) = 2 ≠ 1.
Condition C3 fails (limit ≠ function value) → discontinuous (removable type — could be fixed by setting f(0)=1).
Q7. f(x) = |x|/x for x≠0 and f(0)=0. At x=0, f has:
🤯 T4. f(x) = x sin(1/x) for x≠0, f(0)=0. Is f continuous at x=0?
|sin(1/x)| ≤ 1 for all x≠0 → |x sin(1/x)| ≤ |x| → 0.
By squeeze theorem: limx→0 x sin(1/x) = 0 = f(0).
All three conditions satisfied → f is continuous at x=0. Trap: sin(1/x) oscillates wildly near 0, so students assume discontinuity. The x factor “squeezes” the function to 0. This is the classic squeeze theorem application.
🤯 T5. f(x) = k(x²+2) for x ≤ 0, and f(x) = 3x+1 for x > 0. Find k for continuity.
LHL at x=0: limx→0− k(x²+2) = k(0+2) = 2k.
RHL at x=0: limx→0+ (3x+1) = 1.
f(0) = k(0+2) = 2k (using the x≤0 piece).
Continuity: 2k = 1 → k = 1/2. Always identify which piece gives f(a) when x=a is the boundary point — here x=0 belongs to the first piece (x≤0).
4. Differentiability at a Point
4.1
Definition: First Principles (Limit of Difference Quotient)
LHD and RHD must both exist and be equal for differentiability
A function f(x) is differentiable at x = a if the following limit exists (is finite):
⚡ Derivative by First Principles
f’(a) = lim[h→0] [f(a+h) − f(a)] / h (if this limit exists)
Left-Hand Derivative (LHD):
Lf’(a) = lim[h→0⁺] [f(a−h) − f(a)] / (−h)
= lim[h→0⁺] [f(a) − f(a−h)] / h
Right-Hand Derivative (RHD):
Rf’(a) = lim[h→0⁺] [f(a+h) − f(a)] / h
Condition for differentiability at x = a:
LHD = RHD = finite number
(Both must exist and be equal)
Geometrically: f is differentiable at x = a iff the graph has a unique tangent
(no corner, no cusp, no vertical tangent) at that point.
If LHD ≠ RHD, the function has a corner at x = a — it is continuous but NOT differentiable. The classic example is |x| at x = 0.
Worked Example — Differentiability of |x| at x = 0
Show that f(x) = |x| is not differentiable at x = 0.
But |x| IS continuous at x = 0 (lim = 0 = f(0)). ← Classic NDA example.
Fig 3: Left — smooth curve (unique tangent, differentiable). Middle — corner at |x| (two slopes, not differentiable). Right — cusp (infinite slope, not differentiable).
5. Relationship: Continuity vs Differentiability
5.1
The Fundamental Implication Chain
Differentiability is a STRONGER condition than continuity
Differentiable at a
⇒
Continuous at a
BUT
Continuous at a
⇏
Differentiable at a
⚡ Implication Rules — Memorise the Direction
RULE 1 (Forward): Differentiable ⇒ Continuous
If f is differentiable at x=a, then f MUST be continuous at x=a.
(Differentiability is the stronger/more demanding condition.)
RULE 2 (NOT Reverse): Continuous ⇏ Differentiable
A continuous function may NOT be differentiable.
Classic example: f(x) = |x| at x=0
— Continuous at x=0 (limit = f(0) = 0) ✓
— NOT differentiable at x=0 (LHD=−1, RHD=+1, corner) ✗
RULE 3 (Contrapositive of Rule 1):
NOT continuous ⇒ NOT differentiable
If f is discontinuous at x=a, it is definitely not differentiable there.
NDA tests this with statement-based questions: “If f is continuous at a, it is differentiable at a.” This is FALSE. The converse of a true implication is not always true.
Function
Continuous at x=0?
Differentiable at x=0?
Reason
|x|
Yes ✓
No ✗
LHD=−1, RHD=+1 (corner)
x|x|
Yes ✓
Yes ✓
LHD=RHD=0 (smooth cusp)
x2/3
Yes ✓
No ✗
Vertical tangent at x=0 (∞ slope)
sin(1/x), f(0)=0
No ✗
No ✗
Oscillates infinitely
x sin(1/x), f(0)=0
Yes ✓
No ✗
Continuous, LHD/RHD don’t exist
Polynomials
Yes everywhere ✓
Yes everywhere ✓
Smooth, no corners
📝 TOPIC-WISE PYQ
Differentiability — NDA-Pattern Questions
Q8. Find the derivative of f(x) = x² at x = 3 using first principles.
(a) Continuous and differentiable (b) Continuous but not differentiable (c) Not continuous (d) Not defined
Answer: (b) Continuous but not differentiable
Continuous: limx→1|x−1| = 0 = f(1) ✓.
LHD at x=1: limh→0+[|1−h−1|−0]/(−h) = lim h/(−h) = −1.
RHD at x=1: limh→0+[|1+h−1|−0]/h = lim h/h = +1.
LHD≠RHD → not differentiable at x=1.
Q10. If f(x) is differentiable at x = a, which of the following is necessarily true?
(a) f is not continuous at x=a (b) f is continuous at x=a (c) f’(a) = 0 (d) f(a) = 0
Answer: (b) f is continuous at x=a
Differentiability ⇒ Continuity (always). This is the forward implication rule. Options (c) and (d) are not necessarily true.
Q11. f(x) = x for x ≤ 1, and f(x) = x² for x > 1. At x=1, f is:
(a) Both continuous and differentiable (b) Continuous but not differentiable (c) Differentiable but not continuous (d) Neither
Answer: (b) Continuous but not differentiable
Continuity: LHL=1, RHL=1, f(1)=1. All equal → continuous ✓.
LHD at x=1: derivative of x = 1. RHD: derivative of x² at x=1 = 2(1) = 2.
LHD=1 ≠ RHD=2 → not differentiable ✗.
🔥 TRICKY QUESTIONS
Differentiability — Corner & Implication Traps
🤯 T6. Show f(x) = x|x| is differentiable at x=0. Find f’(0).
For x≥0: f(x) = x⋅x = x². For x<0: f(x) = x⋅(−x) = −x².
LHD: limh→0+[f(0−h)−f(0)]/(−h) = lim[−h²−0]/(−h) = lim h = 0.
RHD: limh→0+[f(0+h)−f(0)]/h = lim[h²−0]/h = lim h = 0.
LHD = RHD = 0 → f’(0) = 0. Differentiable. Contrast with |x|: f(x)=|x| has LHD=−1, RHD=+1. But f(x)=x|x| “smooths out” the corner.
🤯 T7. “If f(x) is not differentiable at x=a, then f(x) is not continuous at x=a.” Is this TRUE or FALSE?
FALSE — this is the CONVERSE of the true statement, not the contrapositive.
True statement: Differentiable ⇒ Continuous.
Contrapositive (also true): Not continuous ⇒ Not differentiable.
Converse (NOT necessarily true): Continuous ⇒ Differentiable (FALSE — |x| is a counterexample).
Inverse (NOT necessarily true): Not differentiable ⇒ Not continuous (FALSE — |x| at x=0 is continuous but not differentiable). This is a direct NDA statement-type question. Know the four forms: statement, converse, inverse, contrapositive.
📝 Master Formula Sheet — MN11 Limits, Continuity & Differentiability
All critical formulae for rapid pre-exam revision.
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