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Limits Continuity and Differentiability · TOPIC LESSON

Continuity and Differentiability

Nearby behavior, matching values and matching slopes: a rigorous route through limits

JEE Main · JEE Advanced100 min study guide14 indexed sections

BY THE END, YOU CAN

  • Distinguish a function value, a finite limit, an infinite limit and one-sided behavior.
  • Evaluate limits by lawful algebra, standard limits and squeeze arguments.
  • Test continuity and differentiability at ordinary points and piecewise joins.
  • Apply continuity theorems with their interval hypotheses and recognize their limitations.

BEFORE YOU START

  • Function domains, inequalities, absolute values and elementary graphs.
  • Factoring, rationalization and radians.

Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.

CONNECT THE CONCEPT TO THE EXAM

Selected exam questions & worked solutions

These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.

No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.

01 / UNDERSTAND & APPLY

1. Three questions that must be kept separate

At a point a, ask separately whether f(a) is defined, whether f(x) approaches a finite value as x approaches a, and whether the difference quotient approaches a finite slope. These are questions about value, limit and derivative. A function can have a limit while its value is missing or different, and it can be continuous while lacking a derivative.

For example, define f(x)=x+2 when x≠1 and f(1)=8. The nearby values approach 3 at x=1, but the assigned value is 8, so the function is not continuous there. Changing only f(1) to 3 repairs continuity. By contrast, no single assigned value repairs a jump with unequal left and right limits.

Limits are local. Values far from a do not affect lim_{x→a}f(x), and changing the value exactly at a does not affect that limit. The derivative, however, uses f(a) in its numerator, so changing the point value can destroy differentiability even if the surrounding formula is unchanged.

Throughout this lesson, an ordinary derivative at an interior point means a finite two-sided derivative. At endpoints, continuity is understood relative to the domain, and a one-sided derivative may be discussed explicitly. An unbounded slope is not a finite derivative.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

Continuity and differentiability are distinct testsLeft panel: f(x)=x+2 for x not equal to 1, and f(1)=8, plotted for −1≤x≤3. An open circle at (1,3) marks the limiting value and a filled circle at (1,8) marks the actual value. Right panel: y=|x| for −2≤x≤2 is continuous at zero but has left slope −1 and right slope 1. Both panels use exact Cartesian coordinates with different stated visual scales.A matching value and a matching slope are separate testsLimit 3, assigned value 8Continuous at 0, unequal side slopesxy(1,3) open(1,8) filled1xyy = |x|slope −1slope +1Left: redefining f(1)=3 repairs continuity. Right: continuity alone does not remove a corner.
Exact examples: a removable value mismatch and a continuous corner. Open circles exclude a point; filled circles include it.

02 / UNDERSTAND & APPLY

2. What a finite limit actually promises

The statement lim_{x→a}f(x)=L means that f(x) can be forced arbitrarily close to L by keeping x sufficiently close to a while excluding x=a and staying in the domain. The approach point must be an accumulation point of the domain so that there are genuinely nearby allowed inputs.

The epsilon-delta definition states: for every ε>0 there exists δ>0 such that 0<|x−a|<δ implies |f(x)−L|<ε for all allowed x. The output tolerance ε is chosen first; δ may depend on ε but must work for every input in the specified punctured neighborhood.

This definition does not require the function to attain L, and it does not require f(a) to exist. Conversely, a function may equal L at many points and still fail to tend to L because other nearby inputs behave differently. A finite table or graph suggests a limit but cannot by itself control every sufficiently close input.

A finite limit is unique. If two different numbers were both limits, choose disjoint small neighborhoods around them. Sufficiently nearby function values would then have to lie in both neighborhoods simultaneously, which is impossible.

ORIGINAL WORKED EXAMPLE

Build a delta from the requested tolerance

Prove lim_{x→2}(3x−1)=5 directly from the definition.

  1. We need |(3x−1)−5|<ε whenever x is sufficiently close to 2.
  2. The left side equals 3|x−2|. For a given ε>0 choose δ=ε/3.
  3. If 0<|x−2|<δ, then |(3x−1)−5|=3|x−2|<3δ=ε.
  4. This works for every positive ε and every qualifying input, completing the proof. There is no requirement that δ be the largest possible choice.

Answer The choice δ=ε/3 proves the limit is 5.

03 / UNDERSTAND & APPLY

3. One-sided behavior identifies jumps and poles

The left limit uses x<a and the right limit uses x>a. At an interior accumulation point approached from both sides, a finite two-sided limit exists exactly when both one-sided limits exist and are equal. Equal-looking formulas are not required; equal limiting values are.

For |x|/x, the right-hand value is 1 and the left-hand value is −1, so the two-sided limit at zero does not exist. Defining the function at zero cannot change this disagreement. For 1/x, the right limit is +∞ and the left limit is −∞, again with no common two-sided limit.

The statement f(x)→+∞ means that for every positive threshold M, the function eventually exceeds M near the approach point. It describes unbounded growth, not convergence to a real number called infinity. Both sides of 1/x² tend to +∞ at zero, so an extended infinite-limit statement is valid, but no finite limit exists.

A limit at infinity concerns inputs leaving every bounded interval. It is distinct from an infinite limit near a finite point. A function may approach a horizontal asymptote while crossing it repeatedly; an asymptote is a limiting description, not a barrier that the graph is forbidden to cross.

ORIGINAL WORKED EXAMPLE

An absolute value makes the two sides different

Evaluate the one-sided limits of (x²−1)/|x−1| at x=1.

  1. Factor x²−1=(x−1)(x+1). For x>1, |x−1|=x−1 and the expression is x+1.
  2. The right-hand limit is therefore 2.
  3. For x<1, |x−1|=−(x−1), so the expression is −(x+1). The left-hand limit is −2.
  4. The finite one-sided limits are unequal, so the two-sided limit does not exist. Canceling x−1 against |x−1| without its sign would give a false answer.

Answer Left limit −2; right limit 2; no two-sided limit.

04 / UNDERSTAND & APPLY

4. Indeterminate forms call for structure, not arithmetic

If f→L and g→M with finite limits, sums, differences and products tend to the corresponding combinations of L and M. A quotient tends to L/M only when M≠0. Composition with a function continuous at L permits substitution into that outer function.

The symbols 0/0 and ∞/∞ are indeterminate forms, not numerical answers. They identify a situation where the simplest quotient law does not decide the limit. Different functions producing the same formal shape can have different limits or no limit at all.

Factoring and canceling a common nonzero factor is valid on a punctured neighborhood even when it is invalid at the approach point itself. That is enough for a limit calculation because the limit excludes the point. Rationalizing a difference of square roots is another way to expose such a factor.

After a transformation, retain the local domain. A square root requires a nonnegative radicand and a denominator must remain nonzero. For limits at infinity, dividing by a dominant power is useful, but √(x²)=|x|, not x on both tails.

ORIGINAL WORKED EXAMPLE

Rationalization removes a zero-over-zero form

Find lim_{x→0}(√(4+x)−2)/x.

  1. Near zero the radicand is positive. Direct substitution produces 0/0, so it does not evaluate the limit.
  2. Multiply numerator and denominator by √(4+x)+2. For x≠0 the expression becomes x/[x(√(4+x)+2)].
  3. Cancel x on the punctured neighborhood to obtain 1/(√(4+x)+2).
  4. This new expression is continuous at zero, giving 1/(2+2)=1/4. The original need not be defined at zero for this limit to exist.

Answer The limit is 1/4.

05 / UNDERSTAND & APPLY

5. Match the small variable and retain its scale

In radians, lim_{u→0}(sin u)/u=1. A geometric squeeze for small positive u gives cos u≤sin u/u≤1, and symmetry handles negative u. This limit underlies the derivative of sine; using that derivative to prove the same foundational limit would be circular in a development that has not established it independently.

Consequences include lim_{u→0}(1−cos u)/u²=1/2, obtained from 1−cos u=2sin²(u/2), and lim_{u→0}(tan u)/u=1. Replacing u by kx requires retaining the factor k. Ratios such as sin(3x)/sin(5x) tend to 3/5, not 1.

Other standard limits are (eᵘ−1)/u→1 and ln(1+u)/u→1 as u→0, with 1+u>0. For a>0, (aᵘ−1)/u→ln a. These can be obtained from the established derivative or integral definitions of exponential and logarithmic functions; state which foundational results are available rather than treating them as algebraic identities.

For a positive expression raised to a variable exponent, take a logarithm of the whole expression and find its limit first. The form 1^∞ is indeterminate. If log y→c, continuity of the exponential gives y→eᶜ. Positivity near the approach point is needed for this method.

ORIGINAL WORKED EXAMPLE

A trigonometric scale and an exponential-type limit

Evaluate lim_{x→0}(1−cos(3x))/x² and lim_{x→0}(1+2x)^(3/x).

  1. For the first limit, write (1−cos(3x))/x²=2[sin(3x/2)/(3x/2)]²·9/4. The bracket tends to 1, giving 9/2.
  2. For the second expression, 1+2x is positive on a neighborhood of zero. Let y=(1+2x)^(3/x).
  3. Then ln y=(3/x)ln(1+2x)=6[ln(1+2x)/(2x)], which tends to 6.
  4. Exponentiating the limiting logarithm gives y→e⁶. Substituting the limiting base 1 before accounting for the diverging exponent would lose the effect.

Answer The limits are 9/2 and e⁶, respectively.

06 / UNDERSTAND & APPLY

6. Oscillation can be suppressed or can destroy a limit

The squeeze theorem states that if g(x)≤f(x)≤h(x) near a and both bounding functions tend to the same finite L, then f(x)→L. The inequalities need only hold sufficiently near the approach point, not necessarily everywhere.

Because |sin(1/x)|≤1, the function x²sin(1/x) tends to zero at zero: its absolute value is at most x². The oscillation becomes increasingly rapid, but its amplitude shrinks to zero. Frequency of oscillation alone does not determine whether a limit exists.

By contrast, sin(1/x) has no limit at zero. Along xₙ=1/(π/2+2πn) it equals 1, while along yₙ=1/(3π/2+2πn) it equals −1; both sequences approach zero through positive inputs. One-sided convergence already fails.

The sequential test says a finite limit L requires f(xₙ)→L for every domain sequence xₙ→a with xₙ≠a. Two sequences with different limiting outputs disprove the limit. One successful sequence cannot prove it, because other approaches may behave differently.

ORIGINAL WORKED EXAMPLE

A vanishing envelope controls every input

Show lim_{x→0}x sin(1/x)=0, but explain why lim_{x→0}sin(1/x) does not exist.

  1. For x≠0, |x sin(1/x)|≤|x|. Since |x|→0, the squeeze theorem gives the first limit as zero.
  2. This proof controls all sufficiently close inputs, not just a selected numerical sequence.
  3. For the unscaled sine, take xₙ=1/(π/2+2πn) and yₙ=1/(3π/2+2πn). Both tend to zero, while their outputs are 1 and −1.
  4. If a limit existed, both output sequences would approach the same value. They do not, so no limit exists.

Answer The scaled limit is 0; the unscaled sine has no limit.

07 / UNDERSTAND & APPLY

7. Dominant terms require the correct sign on each tail

For a rational function, compare the highest powers of x after dividing numerator and denominator by a suitable common power. Equal degrees give the ratio of leading coefficients; a smaller numerator degree gives zero; a larger numerator degree requires further growth and sign analysis rather than a universal “infinity” answer.

Radicals often need rationalization before dominance becomes visible. In particular, √(x²+c) behaves like |x| as |x| grows. The expressions √(x²+c)−x on the positive and negative tails therefore behave very differently.

An asymptotic equivalent can be useful, but cancellation of leading terms demands the next surviving order. Saying both √(x²+1) and x are approximately x does not justify subtracting them and declaring the difference exactly zero. Rationalization exposes the small residual safely.

ORIGINAL WORKED EXAMPLE

The positive and negative tails are different

Find the behavior of √(x²+1)−x as x→+∞ and as x→−∞.

  1. Rationalization gives √(x²+1)−x=1/(√(x²+1)+x), wherever the denominator is nonzero, which it is here.
  2. As x→+∞, the denominator grows without bound, so the difference tends to 0 from above.
  3. As x→−∞, use the original expression: √(x²+1)≥|x|=−x, so √(x²+1)−x≥−2x→+∞.
  4. The two conclusions are compatible because √(x²) is |x|. Replacing it by x on the negative tail would erase the dominant growth.

Answer Limit 0 as x→+∞; divergence to +∞ as x→−∞.

08 / UNDERSTAND & APPLY

8. Continuity matches the nearby limit to the assigned value

A function is continuous at an interior point a when f(a) exists, lim_{x→a}f(x) exists as a finite number, and that limit equals f(a). At an endpoint of an interval, use the one-sided limit from within the domain. More generally continuity is relative to allowed domain inputs.

Polynomials are continuous everywhere. Rational functions are continuous where their denominators do not vanish, and elementary compositions are continuous wherever every stage lies in its proper domain. A denominator zero does not automatically imply an infinite discontinuity; a canceled factor may instead give a removable hole.

A removable discontinuity has a finite limit but a missing or mismatched value. A jump has finite unequal one-sided limits. Infinite behavior and persistent oscillation are other ways continuity can fail. These classifications describe local behavior rather than merely the visual presence of a broken graph.

For a piecewise function, test each join explicitly. First compute the two one-sided limits from their respective formulas, then compare both with the value supplied by the branch containing the join. Matching just the two side limits is insufficient if the assigned point value is different.

ORIGINAL WORKED EXAMPLE

Choose the value that repairs a removable hole

Let f(x)=(x²−9)/(x−3) for x≠3 and f(3)=k. For which k is f continuous at 3?

  1. For x≠3, factor and cancel to get f(x)=x+3.
  2. Therefore the two-sided limit at 3 is 6.
  3. Continuity requires the assigned value f(3)=k to equal that limit, so k=6.
  4. With this value the entire function is x+3, including the previously missing point. Any other k leaves the same limit but creates a discontinuity.

Answer k=6.

09 / UNDERSTAND & APPLY

9. A finite slope is a stronger local requirement

If f is differentiable at a, then f(a+h)−f(a)=h·[f(a+h)−f(a)]/h. The bracket tends to the finite derivative while h→0, so the product tends to zero. Thus f is continuous at a. This proof explains why a jump or a mismatched point value rules out differentiability immediately.

The converse is false. The function |x| is continuous at zero, but its difference quotient there is |h|/h, with left limit −1 and right limit 1. A corner has a perfectly well-defined value and matching nearby values but incompatible local slopes.

At a piecewise join, establish continuity first, then compute one-sided difference quotients using the actual assigned value. If both approach the same finite number, the derivative exists. Differentiating the neighboring formulas and equating their endpoint limits is a useful shortcut only when its hypotheses are justified; it does not replace the definition in every oscillatory case.

Differentiability does not require the derivative function to be continuous. Define f(x)=x²sin(1/x) for x≠0 and f(0)=0. Its derivative at zero is lim h sin(1/h)=0, but for x≠0, f′(x)=2x sin(1/x)−cos(1/x), which has no limit at zero.

10 / UNDERSTAND & APPLY

10. Match values before matching derivatives

Parameter problems often supply enough freedom to make a piecewise function continuous or differentiable. Continuity gives an equation for the matching values. Differentiability adds an equation for the matching one-sided slopes. Solve these in that order so that the derivative calculation uses a consistent point value.

It is possible for continuity to hold for many parameters while differentiability selects only one. It is also possible for no choice to remove a corner or an oscillatory difference quotient. Do not assume that every parameter problem has a solution merely because there are two unknowns and two apparent conditions.

For powers times bounded oscillations at zero, inspect the difference quotient separately from the function itself. A factor strong enough to force f(x)→0 may not be strong enough to force f(x)/x to a finite limit. The derivative loses one power of the small variable in that quotient.

ORIGINAL WORKED EXAMPLE

Two matching conditions determine two parameters

Let f(x)=ax+b for x<1 and f(x)=x² for x≥1. Find a,b so that f is differentiable at 1.

  1. The assigned value is f(1)=1. Continuity requires a+b=1, since the left limit is a+b.
  2. Under that condition, the left difference quotient at 1 is a. The right difference quotient is [(1+h)²−1]/h=2+h for h>0, tending to 2.
  3. Equal one-sided derivatives require a=2. The continuity equation then gives b=−1.
  4. For a=2,b=−1, both values and slopes match. The derivative at the join is 2, though the two formulas differ away from the join.

Answer a=2 and b=−1; f′(1)=2.

11 / UNDERSTAND & APPLY

11. Interval hypotheses turn local continuity into global conclusions

The intermediate value theorem says that a function continuous on [a,b] attains every value between f(a) and f(b). In particular, opposite signs at the endpoints guarantee at least one zero inside. The theorem gives existence, not uniqueness; a separate monotonicity or other argument is needed to count roots.

The extreme value theorem says that a continuous real function on a closed bounded interval attains a maximum and a minimum. Dropping closedness or boundedness can destroy attainment. For example, f(x)=x on (0,1) has neither a maximum nor a minimum, even though it is continuous and bounded.

Continuity of a function on an interval does not force monotonicity, differentiability or an inverse. These are different properties. Likewise, a function can attain every intermediate value without being continuous; the theorem is a sufficient-condition statement, not a characterization.

When proving a root exists for an algebraic or transcendental expression, first choose a closed interval contained in its domain. A sign change across a pole is not enough: 1/x changes sign across zero but has no zero anywhere in its domain.

ORIGINAL WORKED EXAMPLE

A sign change gives existence, a second argument gives uniqueness

Show that x³+x−1=0 has exactly one root in (0,1).

  1. The polynomial is continuous on [0,1]. Its values are −1 at 0 and 1 at 1, so the intermediate value theorem gives at least one root in (0,1).
  2. For y>x, the difference (y³+y−1)−(x³+x−1)=(y−x)(y²+xy+x²+1).
  3. The first factor is positive and the second is positive because y²+xy+x²≥0. Thus the polynomial is strictly increasing on ℝ.
  4. A strictly increasing function has at most one zero. Combining this with existence proves exactly one root in the interval.

Answer Exactly one root lies in (0,1).

12 / UNDERSTAND & APPLY

12. Original practice: limits and nearby behavior

Identify the relevant domain and method before calculating. All trigonometric arguments are in radians.

ORIGINAL PRACTICE · 1

Find lim_{x→2}(x³−8)/(x−2).

Reveal answer & reasoning

12.

For x≠2 the quotient is x²+2x+4. Its limit at 2 is 4+4+4=12.

ORIGINAL PRACTICE · 2

Find lim_{x→0}sin(5x)/(2x).

Reveal answer & reasoning

5/2.

Write the expression as (5/2)[sin(5x)/(5x)] and apply the standard sine limit.

ORIGINAL PRACTICE · 3

Find lim_{x→0}(e^(3x)−1)/x.

Reveal answer & reasoning

3.

Factor 3 times the standard ratio (e^(3x)−1)/(3x), whose limit is 1.

ORIGINAL PRACTICE · 4

Find lim_{x→0}ln(1+4x)/x.

Reveal answer & reasoning

4.

The argument is positive near zero. Write 4[ln(1+4x)/(4x)] and use the logarithm limit.

ORIGINAL PRACTICE · 5

Find lim_{x→∞}(3x²−x+1)/(2x²+5).

Reveal answer & reasoning

3/2.

Divide both numerator and denominator by x². All inverse-power terms tend to zero.

ORIGINAL PRACTICE · 6

Find lim_{x→0}x²cos(1/x).

Reveal answer & reasoning

0.

The absolute value is bounded by x², so the squeeze theorem applies.

ORIGINAL PRACTICE · 7

Does lim_{x→0}|x|/x exist?

Reveal answer & reasoning

No.

The right limit is 1 and left limit is −1; defining the value at zero cannot change them.

ORIGINAL PRACTICE · 8

Find lim_{x→0}(1+x)^(2/x).

Reveal answer & reasoning

e².

The base is positive near zero. Its logarithm is 2ln(1+x)/x→2, so exponentiation yields e².

13 / UNDERSTAND & APPLY

13. Original practice: values, slopes and theorem conditions

At a join, state both the value condition and the derivative condition when relevant. For theorem questions, name the missing or satisfied hypothesis.

ORIGINAL PRACTICE · 1

Let f(x)=sin x/x for x≠0 and f(0)=c. Choose c for continuity.

Reveal answer & reasoning

c=1.

The standard sine limit is 1, so continuity requires the assigned value to match it.

ORIGINAL PRACTICE · 2

Is f(x)=|x−2| continuous and differentiable at 2?

Reveal answer & reasoning

Continuous, but not differentiable.

Both nearby values tend to zero, its value at 2. The left slope is −1 and right slope is 1.

ORIGINAL PRACTICE · 3

Let f(x)=x² for x≤0 and f(x)=ax for x>0. Find a for differentiability at zero.

Reveal answer & reasoning

a=0.

Continuity holds for every a because both side limits are zero. The left derivative is 0 and the right derivative is a, so equality requires a=0.

ORIGINAL PRACTICE · 4

Define f(x)=x sin(1/x) for x≠0 and f(0)=0. Is it differentiable at zero?

Reveal answer & reasoning

No, although it is continuous there.

The function tends to zero by squeeze, but its difference quotient is sin(1/h), which has no limit.

ORIGINAL PRACTICE · 5

Define f(x)=x²sin(1/x) for x≠0 and f(0)=0. Find f′(0).

Reveal answer & reasoning

0.

The difference quotient is h sin(1/h), whose absolute value is at most |h|→0.

ORIGINAL PRACTICE · 6

Does a sign change from 1/(−1) to 1/1 guarantee a root of 1/x in (−1,1)?

Reveal answer & reasoning

No.

The function is not continuous on [−1,1] because it is undefined at zero. The intermediate value theorem does not apply.

ORIGINAL PRACTICE · 7

Must a continuous function on [−2,3] attain its largest value?

Reveal answer & reasoning

Yes.

The interval is closed and bounded and the function is continuous, so the extreme value theorem guarantees attainment.

ORIGINAL PRACTICE · 8

If a function is discontinuous at an interior point, can it have a finite derivative there?

Reveal answer & reasoning

No.

Finite differentiability implies continuity. Its contrapositive rules out differentiability at any discontinuity.

14 / UNDERSTAND & APPLY

14. A hierarchy of tests, not a list of shortcuts

For a limit, examine nearby inputs and the approach direction. For continuity, compare that limit with the assigned point value. For differentiability, divide the local change by the input increment and require one common finite slope. This hierarchy avoids using a later property before an earlier necessary condition has been checked.

Common errors include treating 0/0 as zero, canceling an absolute value without its sign, using √(x²)=x on negative inputs, forgetting radians, declaring an oscillatory limit from a few sampled values, and assuming that continuity implies differentiability.

For a piecewise function, the branch at the join determines the actual value. For a theorem on an interval, poles and missing endpoints matter. For an infinite limit, use language of divergence rather than calling infinity a finite function value.

A strong final check asks what would make the conclusion fail: unequal one-sided limits, an unshrinking oscillation, a mismatched point value, incompatible slopes or a missing theorem hypothesis. Answer that question explicitly and the calculation becomes a complete argument.

KEEP THE BIG IDEAS

Quick revision

  • A limit concerns nearby values and is independent of the assigned value at the approach point.
  • Two-sided limits require compatible one-sided behavior.
  • Continuity matches the finite limit to the actual value.
  • Differentiability implies continuity, but neither continuity nor differentiability implies continuity of the derivative.
  • Existence and attainment theorems require their full interval hypotheses.
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