Why this chapter decides the whole calculus block
Differentiation is defined as a limit. Integration is defined as a limit of sums. Continuity and differentiability are stated in terms of limits. A student who is shaky here does not have one weak chapter - they have an unstable foundation under roughly a third of the mathematics paper.
The good news is that the chapter is finite and highly patterned. There are perhaps eight standard limits, four indeterminate forms that matter, and two theorems. Once those are secure, most questions become recognition rather than invention.
How JEE actually asks Limits
NTA does not publish chapter-wise weightage, so any figure online is a coaching estimate from past papers; check the official NTA information bulletin for the syllabus and pattern that apply to your session.
Past papers show three recurring types. Direct evaluation using standard limits and algebraic manipulation is the most common. Indeterminate forms requiring L'Hopital's rule or series expansion come next. Finally, continuity and differentiability of piecewise-defined functions appear regularly, often asking for the value of a parameter that makes a function continuous or differentiable at a point.
Key concepts, compressed
- ✓A limit describes the value a function approaches, not the value it takes. f(a) may be undefined while the limit at a exists.
- ✓The limit exists only if the left-hand and right-hand limits are equal. Piecewise functions are set up to test exactly this.
- ✓0/0 and infinity/infinity are indeterminate: the answer depends on how the numerator and denominator approach their values, so more work is required.
- ✓L'Hopital's rule replaces the quotient by the quotient of derivatives, and may be applied repeatedly while the form stays indeterminate.
- ✓Series expansion is often faster than repeated differentiation, especially for expressions involving sin, cos or e^x near zero.
- ✓Continuity at a requires LHL = RHL = f(a). Differentiability additionally requires the left and right derivatives to match.
Standard results you need before attempting the questions
| Limit | Value | Note |
|---|---|---|
| lim(x→0) sin x / x | 1 | radians only |
| lim(x→0) tan x / x | 1 | |
| lim(x→0) (1 − cos x)/x² | 1/2 | |
| lim(x→0) (e^x − 1)/x | 1 | |
| lim(x→0) (a^x − 1)/x | ln a | |
| lim(x→0) ln(1 + x)/x | 1 | |
| lim(x→0) (1 + x)^(1/x) | e | |
| lim(x→∞) (1 + 1/x)^x | e | |
| lim(x→0) (x − sin x)/x³ | 1/6 | from the series for sin x |
| L'Hopital's rule | lim f/g = lim f′/g′ | 0/0 or ∞/∞ only |
| Continuity at a | LHL = RHL = f(a) |
The five mistakes that cost the most marks
- ✓Applying L'Hopital's rule without checking the form. It is valid for 0/0 and infinity/infinity only; on a form like 2/3 it produces nonsense.
- ✓Using sin x / x → 1 with x in degrees. The standard limits hold in radians, which is why calculus is done in radians throughout.
- ✓Assuming a continuous function is differentiable. |x| is continuous everywhere and has no derivative at zero.
- ✓Ignoring one-sided limits at a piecewise boundary. The limit exists only if both sides agree, and that is usually the whole question.
- ✓Cancelling a factor without noting the domain. (x² − 4)/(x − 2) equals x + 2 for the limit, but the original function is still undefined at x = 2.
Practice set 1: direct evaluation and standard limits
1. Evaluate lim(x→2) (x² − 4)/(x − 2).
4. Substituting gives 0/0, so factorise: (x² − 4)/(x − 2) = (x − 2)(x + 2)/(x − 2) = x + 2 for x ≠ 2. The limit is therefore 2 + 2 = 4. Note the function itself is undefined at x = 2 - the limit describes what it approaches, not what it equals, which is the definition being tested.
2. Evaluate lim(x→0) sin x / x.
1. This is the foundational trigonometric limit, and it holds only when x is in radians. It follows from the squeeze theorem applied to cos x < sin x / x < 1. Nearly every trigonometric limit in the chapter reduces to this one after manipulation.
3. Evaluate lim(x→0) (1 − cos x)/x².
1/2. Use 1 − cos x = 2sin²(x/2), so the expression becomes 2sin²(x/2)/x² = (1/2)(sin(x/2)/(x/2))², which tends to (1/2)(1)² = 1/2. Answering 1 is the common error, from assuming every standard limit evaluates to one.
4. Evaluate lim(x→0) tan x / x.
1. Write tan x / x = (sin x / x)(1/cos x). The first factor tends to 1 and the second tends to 1/cos 0 = 1, so the product is 1. Splitting into known factors like this is the general method and is safer than trying to recall a longer list of results.
5. Evaluate lim(x→0) (e^x − 1)/x.
1. This is a standard limit and is also, by definition, the derivative of e^x at x = 0 - which is 1 because e^x is its own derivative. Recognising standard limits as derivatives at a point is a useful cross-check when you cannot recall the value.
6. Evaluate lim(x→0) ln(1 + x)/x.
1. Another standard result, and again it is a derivative in disguise: it is the derivative of ln(1 + x) at x = 0, which is 1/(1 + 0) = 1. Together with question 5 it makes exponential and logarithmic limits routine.
7. Evaluate lim(x→∞) (1 + 1/x)^x.
e. This is the definition of e, and the form is 1 to the power infinity - indeterminate, which is why the answer is not simply 1. The companion result lim(x→0)(1 + x)^(1/x) = e is the same statement with the substitution x → 1/x, and questions use both.
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Practice set 2: indeterminate forms
8. Why is 0/0 called indeterminate?
Because the value depends on how the numerator and denominator approach zero, not merely on the fact that both do. x/x tends to 1, x²/x tends to 0, and x/x² tends to infinity - all three are 0/0 forms with different answers. That is precisely why more work is required rather than direct substitution.
9. Evaluate lim(x→0) sin 3x / sin 5x.
3/5. Rewrite as (sin 3x / 3x)(3x) divided by (sin 5x / 5x)(5x). Both bracketed factors tend to 1, leaving 3x/5x = 3/5. The general pattern: for lim(x→0) sin(ax)/sin(bx) the answer is a/b, and manufacturing the matching denominator is what makes it work.
10. Evaluate lim(x→0) (1 − cos 2x)/x².
2. Use the identity 1 − cos 2x = 2sin²x, so the expression is 2sin²x/x² = 2(sin x/x)², which tends to 2. Compare with question 3, where the answer was 1/2 - the doubled angle changes the result by a factor of four, which is exactly what the pairing tests.
11. Evaluate lim(x→∞) (3x² + 2x)/(5x² − 1).
3/5. Divide numerator and denominator by the highest power, x²: (3 + 2/x)/(5 − 1/x²), and as x tends to infinity the small terms vanish, leaving 3/5. For rational functions, the limit at infinity is the ratio of leading coefficients when the degrees are equal, zero when the numerator degree is lower, and infinite when it is higher.
12. Evaluate lim(x→0) (√(1 + x) − 1)/x.
1/2. Rationalise by multiplying above and below by (√(1 + x) + 1): the numerator becomes (1 + x) − 1 = x, giving x/(x(√(1 + x) + 1)) = 1/(√(1 + x) + 1), which tends to 1/2. Rationalising is the standard move whenever a surd appears in a 0/0 form.
13. Evaluate lim(x→0) (1 + x)^(1/x).
e. This is the 1 to the power infinity form and the companion to question 7. The general result is that if f(x) tends to 1 and g(x) tends to infinity, then f(x)^g(x) tends to e raised to the limit of g(x)(f(x) − 1). Here that gives e^1 = e.
14. Evaluate lim(x→0) (a^x − 1)/x.
ln a. Write a^x = e^(x ln a), so the expression becomes (e^(x ln a) − 1)/x. Multiplying and dividing by ln a turns it into ln a times a standard limit equal to 1. Setting a = e recovers question 5, since ln e = 1 - a useful consistency check.
Practice set 3: L'Hopital's rule
15. When may L'Hopital's rule be applied?
Only when direct substitution gives 0/0 or infinity/infinity, and when both functions are differentiable near the point with a non-zero denominator derivative. Other indeterminate forms - 0 times infinity, infinity minus infinity, 1 to the power infinity - must first be algebraically rearranged into a quotient of the right type.
16. Evaluate lim(x→0) (x − sin x)/x³.
1/6. Substituting gives 0/0. Applying L'Hopital's rule three times: (1 − cos x)/3x², then sin x/6x, then cos x/6, which tends to 1/6. The series route is faster: sin x = x − x³/6 + …, so x − sin x = x³/6 + …, and dividing by x³ gives 1/6 immediately. Learn both - series is quicker, L'Hopital is more general.
17. What goes wrong if L'Hopital's rule is applied to lim(x→1) (x + 1)/(x + 2)?
You get a wrong answer. Substituting gives 2/3, which is a perfectly determinate value, so the limit is simply 2/3. Differentiating both parts would give 1/1 = 1, which is incorrect. This is why the form must be checked before the rule is used, every single time.
18. Evaluate lim(x→∞) x/e^x.
0. The form is infinity/infinity, so L'Hopital gives 1/e^x, which tends to 0. The general principle is worth internalising: exponential growth beats any polynomial, so x^n/e^x tends to zero for every n, and applying the rule n times makes that explicit.
19. How do you handle a limit of the form 1 to the power infinity using L'Hopital's rule?
Take logarithms first. Set y equal to the expression, take ln of both sides to obtain a product, rearrange that product into a quotient of the 0/0 or infinity/infinity type, apply the rule, then exponentiate the result. Applying L'Hopital directly to a power form is not valid and is a common source of lost marks.
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Practice set 4: continuity
20. What are the conditions for a function to be continuous at x = a?
Three: f(a) must be defined, the limit as x tends to a must exist (so the left- and right-hand limits are equal), and that limit must equal f(a). Failure of any one gives a discontinuity, and questions are usually built so that exactly one fails.
21. For what value of k is f(x) = kx² for x ≤ 1 and f(x) = 4x − 2 for x > 1 continuous at x = 1?
k = 2. The left-hand limit is k(1)² = k and the value f(1) = k. The right-hand limit is 4(1) − 2 = 2. Continuity requires these to be equal, so k = 2. Parameter questions of this shape are the most common continuity item in the paper, and they are answered purely by equating the two one-sided limits.
22. What kind of discontinuity does (x² − 4)/(x − 2) have at x = 2?
A removable discontinuity. The limit exists and equals 4, but the function is undefined there because the denominator vanishes. Defining f(2) = 4 would make it continuous, which is what removable means. Compare this with a jump, where the limit does not exist at all and no redefinition can help.
23. What is a jump discontinuity?
One where the left- and right-hand limits both exist but are unequal, so the limit does not exist and the graph steps abruptly. The greatest-integer function has a jump at every integer. Unlike a removable discontinuity, no choice of function value at the point can repair it.
24. Is f(x) = |x| continuous at x = 0?
Yes. The left-hand limit, the right-hand limit and f(0) are all zero, so all three continuity conditions hold. This matters because |x| is nonetheless not differentiable at zero - it is the standard demonstration that continuity does not imply differentiability, which is the next set's subject.
25. Where is the greatest-integer function discontinuous?
At every integer. Approaching an integer n from the left gives n − 1 while approaching from the right gives n, so the one-sided limits differ and the function jumps by one. Between consecutive integers it is constant and therefore continuous. This function is a favourite because it makes one-sided reasoning unavoidable.
Practice set 5: differentiability
26. What is the relationship between continuity and differentiability?
Differentiability implies continuity, but not the reverse. If a function has a derivative at a point it must be continuous there; a function can be continuous and still have no derivative, as at a corner. Questions frequently offer the converse as a distractor, and it is false.
27. Why is f(x) = |x| not differentiable at x = 0?
Because the left- and right-hand derivatives differ. Approaching from the left the slope is −1 and from the right it is +1, so no single value exists. The graph has a corner there. This is the canonical example, and any function whose graph has a sharp corner fails differentiability at that corner for the same reason.
28. What condition must hold for a piecewise function to be differentiable at a boundary point?
Both continuity and matching one-sided derivatives. Continuity is necessary but not sufficient, so a two-parameter question typically requires two equations: one from equating the function values, another from equating the derivatives. Solving only the first is the standard incomplete answer.
29. Is f(x) = x² sin(1/x) for x ≠ 0, with f(0) = 0, differentiable at x = 0?
Yes. The derivative at zero is the limit of x² sin(1/x)/x = x sin(1/x), and since sin(1/x) is bounded between −1 and 1 while x tends to zero, the product tends to zero by the squeeze theorem. So f′(0) = 0 despite the wild oscillation. Note that f′ itself is not continuous at zero - differentiability does not require a continuous derivative.
30. What is the difference between a corner and a cusp?
At a corner the one-sided derivatives are finite but unequal, as with |x| at zero. At a cusp they are infinite with opposite signs, as with x^(2/3) at zero, where the tangent becomes vertical. Both make the function non-differentiable while continuous, and the exam distinguishes them by asking about the behaviour of the derivative rather than the function.
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How to study this chapter efficiently
- ✓Substitute first, every time. It costs seconds and tells you whether any technique is needed at all.
- ✓Memorise the eight standard limits as a block and note that they hold in radians. Most trigonometric and exponential questions reduce to one of them.
- ✓Learn the series for sin x, cos x and e^x to three terms. They often beat repeated L'Hopital, and they make results like 1/6 in question 16 obvious.
- ✓For any power form, take logarithms before doing anything else.
- ✓For piecewise functions, always write LHL, RHL and f(a) as three separate lines. The answer is then visible rather than reasoned.
- ✓Study Limits immediately before Differentiation. The derivative is defined as a limit, so the two are one topic split across two chapters.
Turn this into active practice
The skill this chapter needs is classification: seeing an expression and knowing within a few seconds which of five or six routes applies. That is recognition, and recognition only develops against varied questions where the method is not signposted by the section heading.
The JEE Limits quiz on QUFF generates fresh questions across standard limits, indeterminate forms, L'Hopital's rule, continuity and differentiability, marks them instantly and explains each answer. Do a mixed set rather than one type at a time, and note for each error whether you chose the wrong method or executed the right one badly.
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