Traps in Limits, Continuity & Differentiability

6 mistake patterns students fall for. 2 high-frequency traps appear in almost every exam.

Applying L'Hopital's rule without indeterminate form

Very CommonFORMULA

L'Hopital's rule is only valid when the limit is in 0/0 or infinity/infinity form. Applying it to other forms gives incorrect results.

Why: Students apply L'Hopital mechanically as a shortcut without first checking whether the indeterminate form condition is satisfied.

WRONG: lim(x->1) (x^2 + 1)/(x + 1) = lim(x->1) 2x/1 = 2 (applied L'Hopital even though direct substitution gives 2/2 = 1)

RIGHT: Direct substitution: (1 + 1)/(1 + 1) = 2/2 = 1. Not 0/0, so L'Hopital does not apply. Just substitute directly.

See pattern: Evaluate Limit Using L'Hopital's Rule →

Confusing continuity with differentiability

Very CommonCASE MISS

A function can be continuous at a point but not differentiable there. Differentiability is a stronger condition than continuity.

Why: Students assume that if a function is continuous, it must also be differentiable. The classic counterexample f(x) = |x| at x = 0 is often forgotten.

WRONG: f(x) = |x| is continuous at x = 0, so it must be differentiable at x = 0.

RIGHT: f(x) = |x| is continuous at x = 0 but NOT differentiable there. LHD = -1 and RHD = +1 are not equal. Differentiability implies continuity, but continuity does NOT imply differentiability.

See pattern: Check Differentiability at a Point →

Wrong standard limit substitution

CommonFORMULA

Misapplying standard limit formulas by not matching the argument correctly, such as using lim(sin x/x) = 1 when the arguments differ.

Why: Students remember the result lim(sin x/x) = 1 but forget that the argument inside sin must match the denominator exactly.

WRONG: lim(x->0) sin(2x)/x = 1 (incorrectly treating sin(2x)/x as sin(x)/x form)

RIGHT: lim(x->0) sin(2x)/x = lim(x->0) [sin(2x)/(2x)] * 2 = 1 * 2 = 2. Always ensure the argument of sin matches the denominator.

See pattern: Evaluate Limit Using Standard Forms →

Forgetting to check both left and right limits

CommonDOMAIN

A limit exists only if both the left-hand limit and right-hand limit exist and are equal. Checking only one side can give wrong conclusions.

Why: Students compute one-sided limit and assume the overall limit equals it, skipping the other side especially for piecewise functions.

WRONG: For f(x) = {x for x < 0, x+1 for x >= 0}: concluding lim(x->0) f(x) = 0 from the left side alone

RIGHT: LHL = lim(x->0-) x = 0, RHL = lim(x->0+) (x+1) = 1. Since LHL is not equal to RHL, the limit does not exist at x = 0.

See pattern: Check Continuity at a Point →

Incorrect expansion terms in Taylor series

CommonFORMULA

Using wrong coefficients or insufficient terms in Taylor/Maclaurin expansion, leading to wrong limits after cancellation.

Why: Students memorize expansions partially or mix up factorial denominators. Stopping the expansion too early causes terms to vanish incorrectly.

WRONG: Using sin x = x - x^2/2 (wrong coefficient; confusing with cos x expansion)

RIGHT: sin x = x - x^3/3! + x^5/5! - ... and cos x = 1 - x^2/2! + x^4/4! - ... Keep enough terms so that cancellation in the numerator still leaves a nonzero leading term.

See pattern: Find Limit Using Taylor/Maclaurin Expansion →

Not verifying conditions of Rolle's/LMVT

OccasionalCASE MISS

Applying Rolle's theorem or LMVT without checking all the required conditions (continuity, differentiability, equal endpoint values for Rolle's).

Why: Students jump to finding c from f'(c) = 0 or f'(c) = [f(b)-f(a)]/(b-a) without verifying the hypotheses of the theorem.

WRONG: Applying Rolle's theorem to f(x) = |x| on [-1, 1] without checking differentiability at x = 0

RIGHT: f(x) = |x| satisfies f(-1) = f(1) = 1 and is continuous on [-1,1], but is NOT differentiable at x = 0. So Rolle's theorem does not apply on (-1, 1).

See pattern: Apply Rolle's or LMVT →

Test yourself

Can you spot these traps under time pressure?

Take a timed quiz on Limits, Continuity & Differentiability and see if you avoid the mistakes above.

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