A limit answers one question: what value does f(x) approach as x gets arbitrarily close to some point a, regardless of what f(a) actually is? The function's value at a is irrelevant to the limit — it might not even be defined there.
Think of it this way. You are walking toward a wall in a dark room. The limit is about where you are heading, not whether you actually hit the wall. The function value at a is whether the wall exists at all. These are separate facts.
Formally: lim(x→a) f(x) = L means for every ε > 0, there exists δ > 0 such that 0 < |x - a| < δ implies |f(x) - L| < ε. You will never be asked this formal definition in NDA, but internalising the idea saves you from errors.
Left-hand limit (LHL) is the value approached from the left: lim(x→a⁻) f(x). Right-hand limit (RHL) is from the right: lim(x→a⁺) f(x). The two-sided limit exists if and only if LHL = RHL = L.
Continuity is a stronger condition than the limit existing. A function f is continuous at x = a if all three of the following hold:
f(a) is defined.lim(x→a) f(x) exists (i.e., LHL = RHL).lim(x→a) f(x) = f(a).Here's the analogy that sticks: continuity means you can draw the graph through x = a without lifting your pen. A limit existing only means the two sides are heading to the same destination — the pen might lift at exactly a (removable discontinuity) or the graph might jump (jump discontinuity).
In NDA, the examiners test this topic through three types of questions: evaluating tricky limits using standard results, checking continuity/differentiability of piecewise functions, and composition of functions at singular points. All three appear in recent papers — the questions in this page are drawn from 2025 and 2026 papers.
These are the tools, not formulas to memorise blindly. Each one has a geometric or algebraic reason behind it.
Algebraic standard limit:
This is just polynomial long division: x^m - 1 = (x-1)(x^{m-1} + x^{m-2} + \cdots + 1), so the ratio is the sum of m terms each approaching 1. More generally:
Exponential standard limit:
Special case a = e: limit equals 1. This is the definition of the exponential function's derivative at 0. For any base a, you replace a^u with e^{u \ln a}, expand, and the leading term gives ln a.
Trigonometric standard limits:
Both hold because near 0, sin x ≈ x and tan x ≈ x.
When a direct substitution gives 0/0, factoring is your first move. The form (x^n - 1)/(x^m - 1) at x → 1 is the canonical NDA version:
Proof: divide numerator and denominator each by (x - 1). Numerator becomes m, denominator becomes n. Done.
When the 0/0 form involves square roots, multiply numerator and denominator by the conjugate. This cancels the 0/0 form algebraically. Example: (√(x+9) - 3)/(√(x+16) - 4) at x → 0. Multiply by conjugates of both:
At x = 0: (4+4)/(3+3) = 8/6 = 4/3. Clean and fast.
When substitution gives 0/0 or ∞/∞, differentiate numerator and denominator separately:
Use this when standard limits don't apply directly or when the factoring approach is messy. Do not apply it when the form is not indeterminate — that is the single most common L'Hopital error in exam halls.
For a piecewise function, check continuity at every boundary point (where the formula changes). At each boundary:
f(a) using whichever piece applies at a itself.For differentiability, additionally check that the derivatives from the left and right pieces match at the boundary. A function can be continuous but not differentiable at a point (classic example: |x| at 0).
If g is continuous at a and f is continuous at g(a), then f∘g is continuous at a. This is why compositions of standard functions (polynomials, sin, cos, tan, exp) are continuous wherever they are defined. For f·g products: if both are continuous at a, so is their product.
Near x = 0: \tan(x^2) \approx x^2 and x|x| = x^2 \text{ (for x > 0)}, -x^2 \text{ (for x < 0)}. Products of these still go to 0, and their quotient approaches 0 too — so the derivative at 0 exists.
For lim(x→1) (x^p - 1)/(x^q - 1), the answer is simply p/q. No work needed. The standard limit (x^m - 1)/(x-1) = m means you divide top exponent by bottom exponent.
Micro-example: (x^{n²-1} - 1)/(x^{n+1} - 1) at x→1 = (n²-1)/(n+1) = (n-1)(n+1)/(n+1) = n-1.
Standard method (L'Hopital): differentiate twice, simplify — 45 seconds. This pattern: 8 seconds.
For limits of the form (a^{trig(x)} - 1)/trig(x) at x→0, split into two parts:
(a^{trig} - 1)/trig — this is the exponential standard limit ln a.sin x / tan x = cos x) — evaluate at 0.Multiply the two parts. No L'Hopital needed.
Example: (10^{sin x} - 1)/tan x = [(10^{sin x} - 1)/sin x] × [sin x/tan x] → ln 10 × 1 = ln 10.
Standard method (L'Hopital): chain rule differentiation — 60 seconds. This split: 15 seconds.
When you have (√A - c)/(√B - d) and both go to 0, rationalize numerator and denominator simultaneously with their own conjugates. The x-factor cancels from both top and bottom.
Result template: (√A - c)/(√B - d) → (√A + c)/(√B + d) evaluated at the limit point.
Example: (√(x²+9) - 3)/(√(x²+16) - 4) at x→0 → (3+3)/(4+4) = 6/8 = 3/4. Wait — check by plugging x=0: numerator = √9 - 3 = 0, denominator = √16 - 4 = 0. After rationalising: (√16 + 4)/(√9 + 3) = 8/6 = 4/3 (with f(x) = x²+9 substituted appropriately). Standard method (L'Hopital): 50 seconds. Conjugate shortcut: 12 seconds.
At each boundary of a piecewise function, write two equations immediately:
f(a).For differentiability, add a third equation: left derivative = right derivative.
With k unknowns and k equations from multiple boundary points, solve the system. This converts a conceptual question into a pure algebra question — 3 equations, 3 unknowns, done in under 2 minutes versus 4+ minutes of uncertain reasoning.
If a question asserts differentiability at a point, you know continuity there is automatic. Conversely, if a function is discontinuous at a, cross out any option claiming differentiability at a immediately — differentiability is impossible without continuity.
For |x|-type functions at 0: always continuous (value matches from both sides), never differentiable (slopes are +1 and −1). This kills two wrong options in 3 seconds.
Step count: reasoning from scratch — 6 steps. Using this implication — 1 step of elimination.
When a limits-and-continuity question appears, run this decision tree in your head:
Step 1 — Identify the form. Substitute the limit point directly. If you get a clean finite answer, you are done. If you get 0/0 or ∞/∞, proceed.
Step 2 — Match to a standard limit. Check: does it look like (x^m - 1)/(x^n - 1) at x→1? Exponential (a^u - 1)/u at u→0? Trig sin(u)/u at u→0? If yes, apply the pattern directly. If the exponent or argument is a function of x, substitute u = that function first.
Step 3 — Rationalise if surds are present. Multiply by conjugates. After cancellation, substitute directly.
Step 4 — L'Hopital as last resort. Differentiate top and bottom. Check form again after one application. Rarely need more than two rounds in NDA.
For continuity/differentiability questions: Go straight to the boundary points. Write LHL = RHL = f(boundary), then solve. Check derivatives if differentiability is asked. Do not spend time on interior points — polynomials, sin, tan are continuous everywhere in their natural domain.
Why this question: Tests whether you recognise the (x^m-1)/(x^n-1) pattern at x→1 and can factor n²-1.
Solving path: Direct substitution gives 0/0. Recognise the form: divide top and bottom each by (x-1). Top becomes n²-1, bottom becomes n+1. So the limit is (n²-1)/(n+1) = (n-1)(n+1)/(n+1) = n-1. That is option (c). Total time: under 20 seconds once you see the structure.
Why this question: Tests the exponential-trigonometric limit split — a technique that NDA examiners reuse regularly.
Solving path: Split: (10^{\sin x} - 1)/\tan x = [(10^{\sin x} - 1)/\sin x] \cdot [\sin x / \tan x]. As x→0, \sin x → 0, so the first bracket uses \lim_{u→0}(10^u - 1)/u = \ln 10. The second bracket is \cos x → 1. Product: \ln 10 \cdot 1 = \ln 10. Option (c).
Why this question: Multi-boundary piecewise function requiring you to set up and solve a system — the kind of question where knowing the method is everything.
Solving path (for q): Continuity at x = 3: value from middle piece = 3-1 = 2. Value from right piece = 9p + 3q + 2. Setting equal: 9p + 3q = 0, so q = -3p. Differentiability at x = 3 (f' continuous): derivative of middle piece at 3 is 1. Derivative of right piece at 3 is 6p + q. Setting equal: 6p + q = 1. Substituting q = -3p: 6p - 3p = 1, so p = 1/3. Then q = -3(1/3) = -1. Option (a).
Why this question: Tests continuity and differentiability of a product of two functions near x=0 using approximation — a high-thinking question that becomes easy once you expand near 0.
Solving path: p(x) = \tan(x^2) \cdot x|x|. Near x=0, \tan(x^2) \approx x^2, so p(x) \approx x^2 \cdot x|x| = x^3|x|. At x=0: p(0) = 0. As x→0, p(x)→0 = p(0): continuous. For differentiability: p'(0) = \lim_{x→0} p(x)/x = \lim_{x→0} x^2|x| = 0. Exists. Both statements correct: option (c).
Why this question: Rationalisation of a surd limit — tests algebraic technique speed.
Solving path: f(x) = x^2 + 9. At x=0: numerator = \sqrt{9}-3 = 0, denominator = \sqrt{16}-4 = 0. Rationalise: multiply by (\sqrt{f}+3)(\sqrt{f+7}+4)$ in cross-fashion. After algebra, the ratio simplifies to (\sqrt{f+7}+4)/(\sqrt{f}+3). At x=0: f=9, f+7=16. Ratio = (4+4)/(3+3) = 8/6 = 4/3`. Option (c).
Applying L'Hopital to non-indeterminate forms. If direct substitution gives 3/0 (not 0/0), the limit is infinity or does not exist — L'Hopital does not apply. Check the form first, always.
Confusing the standard limit base. \lim_{u→0}(a^u-1)/u = \ln a, not \log_{10} a. If you write \log_{10} e instead of \ln 10, you have confused natural log with common log — these are different by a factor of \ln 10 \approx 2.303. In the 10^{\sin x}/\tan x question, \ln 10 is correct; \log_{10} e is wrong.
Checking continuity only from one side at a piecewise boundary. At x = a where the formula changes, you must compute LHL, RHL, and f(a) separately. Checking only LHL = f(a) and declaring continuity is wrong — RHL might differ.
Concluding differentiability from continuity. Continuity at a point does not guarantee differentiability. The converse direction (differentiability implies continuity) is the safe one. |x| at 0 is the standard counterexample — continuous but not differentiable.
Forgetting that f(x) for x² < 1 means −1 < x < 1, not just x < 1. The condition x² < 1 covers x \in (-1, 1). This affects which formula applies near x = -1 and x = 1 independently. Misreading the domain flips the continuity analysis entirely.
Skipping the derivative sign at a piecewise boundary. When asked whether f' is continuous at x = a, you must compute derivatives of each piece and check they match at a — it is not enough that f itself is continuous there.