Calculus looks scary, but it all begins with one simple idea: what value does a function approach? In the NDA exam, Limits, Continuity and Differentiability is a high-yield topic that quietly underpins derivatives and integrals too. Get the three definitions clear, memorise a handful of standard limits, and you can grab 4–6 reliable marks every attempt.
Why This Topic Is Worth Your Time
Every NDA Maths paper carries questions on limits and continuity, and most of them are direct, formula-based and fast to solve. Because the same standard limits and tricks repeat year after year, this is one of the most predictable scoring areas in the whole calculus block.
There is a bigger pay-off too. Limits are the foundation on which derivatives and integrals are built, so the time you invest here strengthens three chapters at once. A student who is comfortable spotting a 0/0 form will sail through differentiation later. That is why Cavalier teachers treat this as the true starting line of calculus.
First decide what kind of limit you are facing — a clean substitution, a 0/0 form, an ∞/∞ form, or a standard result in disguise. Choosing the right method is most of the battle; the algebra is the easy part.
The Idea of a Limit
The limit of a function f(x) as x approaches a value a is the single value that f(x) gets closer and closer to as x gets closer to a. We write it as:
limx→a f(x) = L means f(x) → L as x → a.
The limit exists only if the left-hand limit equals the right-hand limit:
limx→a− f(x) = limx→a+ f(x) = L
The crucial subtlety is that the limit does not care about the value of f at a itself — only about the behaviour around a. A function can have a perfectly good limit at a point where it is not even defined. This is exactly why limits let us study tricky points like 0/0.
A limit exists only when the function approaches the same value from both sides. If the left and right limits disagree, the limit simply does not exist.
Methods of Evaluating Limits
When you plug x = a into f(x), three things can happen. If you get a finite number, that is your answer — done. If you get a form like 0/0 or ∞/∞, the limit is indeterminate and needs work. Here are the staple techniques.
Direct substitution
For polynomials and most well-behaved functions, just substitute x = a. For example, limx→2 (x2 + 1) = 5.
Factorisation
If substitution gives 0/0, factor the numerator and denominator, cancel the common factor, then substitute again. This clears most rational-function limits.
Rationalisation
When square roots create the 0/0 form, multiply by the conjugate to remove the surd, then cancel and substitute. For example, in limx→0 (√(1 + x) − 1)/x, multiplying top and bottom by (√(1 + x) + 1) clears the root and leaves 1/(√(1 + x) + 1), which tends to 1/2.
Limits at infinity
When x → ∞, divide every term of a rational function by the highest power of x present. Each term like 1/x or 1/x2 then heads to zero, and only the leading coefficients survive. This is how you handle the ∞/∞ form for polynomials over polynomials without any rule at all.
A 0/0 form is a signal, not a dead end. It tells you there is a common factor hiding — cancel it and the limit reappears. The same goes for ∞/∞: divide through and watch the small terms vanish.
Standard Limits You Must Memorise
A large share of NDA limit questions are just these results in a small disguise. Learn them cold.
limx→0 (sin x)/x = 1 (x in radians)
limx→0 (tan x)/x = 1
limx→0 (1 − cos x)/x2 = 1/2
limx→0 (ex − 1)/x = 1
limx→0 (ax − 1)/x = loge a
limx→0 (log(1 + x))/x = 1
limx→0 (1 + x)1/x = e
limx→a (xn − an)/(x − a) = n·a(n−1)
Almost every trigonometric limit reduces to the (sin x)/x result once you adjust the angle. For example, limx→0 (sin 5x)/x = 5 because you write it as 5 × (sin 5x)/(5x) and the bracket tends to 1. The same multiply-and-divide trick handles (tan 7x)/x → 7 and (sin ax)/(sin bx) → a/b, both of which appear regularly in the paper.
The exponential and logarithm results are just as important and often more neglected by students. Whenever you see ex − 1, log(1 + x), or a base raised to x minus one, near x = 0, reach for these standard limits instead of trying to expand anything. They convert what looks like an advanced question into a one-line answer. Recognising the pattern is faster than any algebra you could do, so train your eye to spot these shapes immediately.
The result (sin x)/x → 1 holds only when x is in radians. If a question is in degrees, convert first or you will get a wrong factor of π/180.
L'Hopital's Rule for 0/0 and ∞/∞
When a limit gives the indeterminate form 0/0 or ∞/∞, you can differentiate the top and bottom separately and try again.
If limx→a f(x)/g(x) is of the form 0/0 or ∞/∞, then
limx→a f(x)/g(x) = limx→a f′(x)/g′(x)
provided the right-hand limit exists. You may repeat the rule if the form is still 0/0.
This is a powerful shortcut for exam conditions because it turns a messy algebraic limit into a quick derivative. For instance, limx→0 (sin x)/x becomes limx→0 (cos x)/1 = 1 in a single line, and limx→0 (ex − 1)/x becomes limx→0 ex/1 = 1 just as easily.
L'Hopital's rule is also a lifesaver when no clever factorisation jumps out at you. If the algebra looks ugly and the form is genuinely 0/0, differentiating top and bottom is usually the calmest path in the exam hall. Just keep an eye on the form after each step: the moment substitution gives a finite number, stop and read off the answer.
Apply L'Hopital's rule only to genuine 0/0 or ∞/∞ forms. Using it on an ordinary limit, or differentiating the whole quotient with the quotient rule, gives a wrong answer.
Continuity at a Point
Informally, a function is continuous if you can draw its graph without lifting your pen. Formally, f is continuous at x = a when its value there equals the limit there.
f(x) is continuous at x = a if all three hold:
- f(a) is defined,
- limx→a f(x) exists (left limit = right limit),
- limx→a f(x) = f(a).
If any one of these fails, the function is discontinuous at that point. A break in the graph, a hole, or a sudden jump all signal discontinuity. A removable discontinuity is a hole where the limit exists but does not equal f(a); a jump discontinuity is where the left and right limits exist but differ.
Polynomials, sine, cosine and the exponential function are continuous everywhere. Rational functions are continuous except where the denominator is zero. Knowing these standard facts lets you spot the only suspicious points instead of checking the whole real line.
For a piecewise function, the only places that can break continuity are the join points where the formula changes. Check the left limit, the right limit and the value there.
Differentiability and Its Link to Continuity
A function is differentiable at x = a if it has a definite slope there — that is, the derivative from the left equals the derivative from the right.
f′(a) = limh→0 [f(a + h) − f(a)] / h, when this limit exists.
Differentiable at a ⇒ Continuous at a.
But Continuous at a does NOT imply Differentiable at a.
The one-way arrow is the heart of this topic. Differentiability is the stronger condition: every differentiable function is continuous, yet many continuous functions fail to be differentiable at sharp corners.
The classic example is f(x) = |x| at x = 0. The graph is unbroken there, so it is continuous, but it has a sharp corner: the left slope is −1 and the right slope is +1. Since the two slopes disagree, the function is not differentiable at x = 0.
To test differentiability of a piecewise function at a join point, you check the left-hand derivative and the right-hand derivative separately and ask whether they match. If they do, the function is smooth there and differentiable; if they differ, you have a corner. Remember to confirm continuity first, because a function that is not even continuous at a point can never be differentiable there. NDA examiners love piecewise functions where you must find a constant that makes the function both continuous and differentiable at once.
Never assume continuity guarantees differentiability. Corners (like |x|) and vertical tangents (like x1/3 at 0) are continuous but not differentiable.
Worked Example: A 0/0 Trigonometric Limit
Let us solve a typical NDA-style limit using a standard result.
Evaluate limx→0 (1 − cos 2x) / x2.
Spotting the identity 1 − cos 2x = 2 sin2x turned a 0/0 form into the standard (sin x)/x result. That single substitution is the move the examiner is testing.
Common Mistakes to Avoid
Most lost marks here come from small oversights rather than hard ideas.
- Forgetting that (sin x)/x → 1 needs radians, not degrees.
- Cancelling a factor without first checking it really is 0/0 — if substitution gives a finite number, you are already done.
- Applying L'Hopital's rule to a limit that is not indeterminate.
- Assuming a continuous function must be differentiable — remember |x|.
- Ignoring the left-hand limit; for one-sided behaviour you must check both sides.
When a limit gives ∞ − ∞ or 0 × ∞, rearrange it into a 0/0 or ∞/∞ form before using L'Hopital. These rules need a fraction.
Previous-Year Style Practice
Try this NDA-pattern question before reading the solution.
Q. For what value of k is the function f(x) = (sin 3x)/x for x ≠ 0 and f(0) = k continuous at x = 0?
Answer: For continuity we need limx→0 f(x) = f(0). Now limx→0 (sin 3x)/x = 3 × limx→0 (sin 3x)/(3x) = 3 × 1 = 3. So for continuity, k = 3.
Notice the standard trick: multiply and divide by the angle inside the sine (here 3x) so the bracket becomes the known (sin θ)/θ → 1 result. This single idea answers a huge share of NDA limit-and-continuity questions.
Quick Revision Before the Exam
Glance over these the night before and the morning of your paper.
- Limit exists ⇔ left-hand limit = right-hand limit.
- 0/0 form? Try factorisation, rationalisation, a standard limit, or L'Hopital's rule.
- Key results: (sin x)/x → 1, (1 − cos x)/x2 → 1/2, (ex − 1)/x → 1, (1 + x)1/x → e.
- Continuity at a: f(a) defined, limit exists, and limit = f(a).
- Differentiable ⇒ Continuous, but NOT the reverse (think |x|).
Drill 10–15 mixed limit problems daily for a week. Speed on standard limits frees up minutes for the heavier derivative and integral questions later in the paper.
Frequently asked questions
What does it mean for a limit to exist?
A limit exists at a point only when the function approaches the same value from both the left and the right. If the left-hand limit and the right-hand limit are different, the limit does not exist, even if the function is defined at that point.
Is the value f(a) needed for the limit at a to exist?
No. A limit depends only on how the function behaves around the point, not at the point itself. A function can have a perfectly valid limit at a place where it is not even defined, which is why limits let us study 0/0 forms.
When can I use L'Hopital's rule in the NDA exam?
Use it only when direct substitution gives the indeterminate form 0/0 or infinity/infinity. You then differentiate the numerator and denominator separately and take the limit again, repeating if the form is still indeterminate.
Does continuity imply differentiability?
No. Differentiability implies continuity, but not the reverse. The function f(x) = |x| is continuous everywhere yet not differentiable at x = 0 because it has a sharp corner where the left and right slopes differ.
Why must x be in radians for the limit of sin x over x?
The result that sin x divided by x tends to 1 as x approaches 0 is derived using radian measure. If x is in degrees, an extra factor of pi over 180 appears, so always convert to radians first.
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