The function f(x) = |x| at x = 0 is:

Continuous but not differentiable

Differentiable but not continuous

Neither continuous nor differentiable

Rolle's theorem does not apply to f(x) = |x| on [−1, 1] because:

f is not differentiable at x = 0

f is not continuous on [−1, 1]

If a function is differentiable at a point, then at that point it is necessarily:

Either — differentiability says nothing about continuity

Bounded but possibly discontinuous

For a piecewise function to be differentiable at a joining point, it is enough that:

Both the values and the one-sided derivatives match there

The two pieces have equal values there

Both the values and the one-sided derivatives match there

Each piece is differentiable on its own domain

The function is bounded near the point

Limits, Continuity and Differentiability

This is the chapter that decides whether calculus feels mechanical or mysterious. Its core idea is a single distinction: a limit asks what a function approaches, which has nothing to do with what the function is at the point. f(x) = (x² − 1)/(x − 1) is undefined at x = 1 and yet its limit there is 2, and grasping why is the whole content of continuity. The chapter then builds a strict hierarchy — differentiable implies continuous implies a limit exists — and almost every question in it turns on knowing that the implications run only one way.

Evaluating Limits

Standard limits worth knowing outright
Limit as x → 0	Value
sin x / x	1 — the foundational one
tan x / x	1
(1 − cos x)/x²	1/2
(eˣ − 1)/x	1
log(1 + x)/x	1
(1 + x)^1/x	e

A limit exists only if the left-hand and right-hand limits are equal. That is why a modulus or a greatest-integer function must always be checked from both sides: lim(x→0) |x|/x is −1 from the left and +1 from the right, so the limit does not exist even though both one-sided limits do. L'Hôpital's rule — differentiate numerator and denominator separately — applies to the indeterminate forms 0/0 and ∞/∞ and to nothing else, so the form must be confirmed before the rule is used.

⚠️ The seven indeterminate forms, and the ones that only look indeterminate

The genuine indeterminate forms are 0/0, ∞/∞, 0 × ∞, ∞ − ∞, 0⁰, 1^∞ and ∞⁰ — and only these need work. Expressions that merely look awkward are often determinate: 0^∞ is 0, ∞^∞ is ∞, and 1/0 is unbounded rather than indeterminate. The distinction matters because applying L'Hôpital to a form that is not 0/0 or ∞/∞ gives a wrong answer confidently. For the exponential forms 1^∞ and 0⁰ the standard route is to take logarithms, converting the problem into 0 × ∞ and then into 0/0, which is why lim (1 + 1/n)ⁿ = e rather than 1.

Continuity

A function is continuous at x = a when three things all hold: f(a) is defined, the limit at a exists, and the two are equal. Failing any one gives a discontinuity, and the type has a name: removable if the limit exists but differs from f(a), a jump if the one-sided limits differ, and infinite if either is unbounded. Questions that give a piecewise function and ask for the value of a constant making it continuous are asking you to set the two one-sided limits equal to f(a).

🎯 Why |x| is continuous everywhere but not differentiable at zero

Continuity asks only that the graph have no break, and the graph of |x| is a single unbroken V — you can draw it without lifting the pen, so it is continuous at 0. Differentiability asks something stronger: that the graph have a well-defined tangent, which means the slope approaching from the left must match the slope approaching from the right. For |x| the left slope is −1 and the right slope is +1, so there is no single tangent at the corner and the derivative does not exist there. This is the cleanest illustration of the hierarchy: differentiable ⟹ continuous, but continuous does not imply differentiable, and every "sharp corner" function is a counterexample to the converse.

The hierarchy, and where each implication fails
Statement	True?	Counterexample if false
Differentiable ⟹ continuous	Yes, always	None exists
Continuous ⟹ differentiable	No	\|x\| at x = 0
Limit exists ⟹ continuous	No	(x²−1)/(x−1) at x = 1, undefined there
Continuous ⟹ limit exists	Yes	None exists
Discontinuous ⟹ not differentiable	Yes — the contrapositive of the first row	None exists

🧠 Check differentiability by matching left and right derivatives, not by differentiating

For a piecewise function the safe test at a joining point is to compute the left-hand derivative and the right-hand derivative from first principles and see whether they agree. Differentiating each piece separately and evaluating at the joint often gives the right answer but silently assumes the derivative is continuous there, which is exactly the thing in question. Two conditions must hold at a joint: the function values must match for continuity, and the derivatives must match for differentiability — and a question asking for two constants a and b is usually asking you to write down one equation from each.

Two named theorems close the chapter and are examined for their hypotheses rather than their conclusions. Rolle's theorem needs f continuous on [a, b], differentiable on (a, b) and f(a) = f(b), and then guarantees a point where f′ = 0. Lagrange's mean value theorem drops the last requirement and guarantees a point where f′ equals the average slope (f(b) − f(a))/(b − a). A question offering |x| on [−1, 1] and asking whether Rolle's theorem applies is testing whether you checked differentiability: the endpoints match and the function is continuous, but the corner at 0 breaks the second hypothesis, so the theorem simply does not apply.

Key takeaways

A limit describes what a function approaches, not its value at the point — the two can differ.
sin x/x → 1, (1 − cos x)/x² → ½, (eˣ − 1)/x → 1 and (1 + x)^1/x → e as x → 0.
A limit exists only if the left and right limits agree — always check both for modulus functions.
L'Hôpital applies only to 0/0 and ∞/∞ — confirm the form before differentiating.
0^∞ is 0, not indeterminate; the seven true indeterminate forms are worth listing once.
Continuity needs f(a) defined, the limit to exist, and the two to be equal.
Differentiable implies continuous, but not conversely — |x| at 0 is the standard counterexample.
Rolle and Lagrange are examined on their hypotheses: check differentiability before applying either.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

lim(x→0) (1 − cos x)/x² equals:
1. 0
2. 1/2
3. 1
4. 2
Show answer
Answer: B — 1/2
Using 1 − cos x = 2sin²(x/2), the expression becomes 2sin²(x/2)/x² = ½ · (sin(x/2)/(x/2))², which tends to ½ · 1 = ½. Two applications of L'Hôpital give the same value. The answer 1 comes from misremembering this as the sin x/x limit, which is a different standard result — the squared denominator is what makes the ½ appear.
The function f(x) = |x| at x = 0 is:
1. Continuous and differentiable
2. Continuous but not differentiable
3. Differentiable but not continuous
4. Neither continuous nor differentiable
Show answer
Answer: B — Continuous but not differentiable
The graph is an unbroken V, so the function is continuous at 0, but the left-hand derivative is −1 and the right-hand derivative is +1, so no single tangent exists and the derivative fails. The third option is impossible for any function, since differentiability always forces continuity — recognising that immediately eliminates it from any option list.
lim(x→0) |x|/x:
1. Equals 1
2. Equals 0
3. Does not exist
4. Equals −1
Show answer
Answer: C — Does not exist
From the right |x| = x so the quotient is +1, and from the left |x| = −x so it is −1. Both one-sided limits exist but they disagree, and a limit requires them to be equal, so the limit does not exist. This is why any expression containing a modulus, a signum or a greatest-integer function must be examined from both sides rather than evaluated directly.
Which of the following is NOT an indeterminate form?
1. 0/0
2. 1^∞
3. 0^∞
4. ∞ − ∞
Show answer
Answer: C — 0^∞
A quantity tending to zero raised to a power tending to infinity is unambiguously zero, so 0^∞ is determinate and needs no special treatment. The other three are genuinely indeterminate: 0/0 and ∞ − ∞ require algebraic work or L'Hôpital, and 1^∞ requires taking logarithms — which is exactly why lim (1 + 1/n)ⁿ is e rather than 1.
For f(x) = (x² − 4)/(x − 2) with f(2) undefined, the discontinuity at x = 2 is:
1. A jump discontinuity
2. Removable
3. An infinite discontinuity
4. Not a discontinuity at all
Show answer
Answer: B — Removable
Cancelling gives x + 2 for every x except 2, so the limit at 2 is 4 and exists, but the function is not defined there — a gap that can be closed simply by defining f(2) = 4, which is what "removable" means. A jump would require the two one-sided limits to differ, and an infinite discontinuity would require the function to blow up, neither of which happens here.
Rolle's theorem does not apply to f(x) = |x| on [−1, 1] because:
1. f(−1) ≠ f(1)
2. f is not differentiable at x = 0
3. f is not continuous on [−1, 1]
4. The interval is not closed
Show answer
Answer: B — f is not differentiable at x = 0
The endpoint values do match, since f(−1) = f(1) = 1, and the function is continuous throughout, so the first and third hypotheses hold. What fails is differentiability on the open interval, because of the corner at x = 0 — and indeed the conclusion fails too, since f′ is never zero anywhere it exists. Questions of this type are testing whether the hypotheses were checked individually rather than assumed.
If a function is differentiable at a point, then at that point it is necessarily:
1. Continuous
2. Not continuous
3. Either — differentiability says nothing about continuity
4. Bounded but possibly discontinuous
Show answer
Answer: A — Continuous
Differentiability is the stronger condition and always implies continuity, because a derivative can only exist where the function has no break. The converse fails, as |x| shows at the origin, so the implication runs strictly one way — which is why "differentiable but not continuous" is never a valid option and can be eliminated on sight.
For a piecewise function to be differentiable at a joining point, it is enough that:
1. The two pieces have equal values there
2. Both the values and the one-sided derivatives match there
3. Each piece is differentiable on its own domain
4. The function is bounded near the point
Show answer
Answer: B — Both the values and the one-sided derivatives match there
Matching values gives continuity, which is necessary but not sufficient; the one-sided derivatives must also agree for a single tangent to exist. Each piece being smooth on its own says nothing about the joint, which is precisely where the failure occurs in examples like |x|. A question giving two unknown constants usually wants one equation from the value match and one from the derivative match.