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Limits and Continuity

A free GRE Mathematics note on limit laws, the squeeze theorem, L'Hopital's rule, the three-condition definition of continuity, and the classic limits every calculus question assumes you know.

Concise answer

Limits are computed by substitution plus the limit laws whenever the pieces behave, and by a named tool — algebraic cancellation, the squeeze theorem, a series expansion, or L'Hopital's rule — exactly when substitution produces an indeterminate form. Continuity at a point is three separate checks, and knowing which check fails classifies the discontinuity.

Definitions

Limit
The value $L$ that $f(x)$ approaches as $x$ approaches $a$; it depends only on values of $f$ near $a$, never on $f(a)$ itself.
Continuity at a point
$f$ is continuous at $a$ when three conditions hold: $f(a)$ is defined, $\lim_{x \to a} f(x)$ exists, and $\lim_{x \to a} f(x) = f(a)$.
Indeterminate form
A symbolic result of substitution, such as $0/0$ or $\infty/\infty$, that does not determine the limit by itself and signals that more work is required.
Removable discontinuity
A discontinuity at $a$ where $\lim_{x \to a} f(x)$ exists but is undefined at $a$ or disagrees with $f(a)$; redefining one point repairs it.

Intuition

A limit is a statement about the neighborhood, not the point: f(a)f(a) can be undefined or wrong and the limit can still exist. That separation is exactly what the third continuity condition tests — whether the point agrees with its neighborhood.

L'Hopital's rule and Taylor expansion are two views of the same fact: near the point, both numerator and denominator are well approximated by their leading polynomial behavior, and the limit is decided by the lowest-order terms that survive.

Concept walkthrough

The limit laws reduce complicated limits to simple ones: limits pass through sums, differences, products, constant multiples, and quotients whose denominator limit is nonzero. For polynomial and rational functions inside their domains this collapses to substitution. The interesting cases are the ones substitution cannot settle — a 0/00/0 or /\infty/\infty form — and the discipline is to name the form before reaching for a tool.

Three tools resolve most indeterminate forms on the exam. Algebraic rewriting (factoring, conjugates) cancels the shared zero directly. The squeeze theorem handles bounded oscillation: if g(x)f(x)h(x)g(x) \le f(x) \le h(x) near the point and gg and hh share a limit LL, then fLf \to L as well — this is how the classic limit limx0(sinx)/x=1\lim_{x \to 0} (\sin x)/x = 1 is established. L'Hopital's rule differentiates numerator and denominator separately, and applies only to genuine 0/00/0 or /\infty/\infty forms; applied to a determinate form it simply gives wrong answers.

Continuity is the pointwise bookkeeping built on limits: f(a)f(a) defined, limit exists, limit equals value. A removable discontinuity fails only the agreement condition, a jump has different one-sided limits, and an infinite discontinuity has an unbounded one-sided limit. GRE Mathematics questions frequently hand you a piecewise function and ask which conditions fail where — answering condition by condition is faster and safer than intuition about the graph.

After this page, you should be able to

  • Evaluate limits with the limit laws and recognize when a nonzero-denominator substitution finishes the problem immediately.
  • State the three-condition definition of continuity at a point and use it to classify discontinuities.
  • Apply the squeeze theorem to bounded-oscillation limits such as $x^2 \sin(1/x)$.
  • Verify an indeterminate form before invoking L'Hopital's rule, and cross-check the result with a series expansion when one is available.

Formulas and assumptions

Classic sine limit

limx0sinxx=1\lim_{x \to 0} \frac{\sin x}{x} = 1

Variables

  • x: real variable measured in radians

Assumptions

  • x is in radians; in degrees the limit is pi/180.
  • Established via the squeeze theorem, so it may be used without L'Hopital's rule.

Squeeze theorem

g(x)f(x)h(x)  near a,  limxag(x)=limxah(x)=L    limxaf(x)=Lg(x) \le f(x) \le h(x) \;\text{near } a,\; \lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L \;\Rightarrow\; \lim_{x \to a} f(x) = L

Variables

  • f: the squeezed function
  • g, h: bounding functions
  • L: the shared limit

Assumptions

  • The bounds hold for all x near a except possibly at a itself.

L'Hopital's rule

limxaf(x)g(x)=limxaf(x)g(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}

Variables

  • f, g: functions differentiable near a
  • a: the limit point (finite or infinite)

Assumptions

  • The original limit is a genuine 0/0 or infinity/infinity indeterminate form.
  • g'(x) is nonzero near a (except possibly at a), and the limit of f'/g' exists or is infinite.

Worked example

One limit, two methods

Evaluate limx01cosxx2\displaystyle \lim_{x \to 0} \frac{1 - \cos x}{x^2} by series expansion, then confirm with L'Hopital's rule.

  1. 1Substitution gives (1cos0)/02=0/0(1 - \cos 0)/0^2 = 0/0, an indeterminate form, so a tool is required.
  2. 2Series method: cosx=1x22!+x44!\cos x = 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \cdots, so 1cosx=x22x424+1 - \cos x = \dfrac{x^2}{2} - \dfrac{x^4}{24} + \cdots (checking the arithmetic: 2!=22! = 2 and 4!=244! = 24).
  3. 3Divide by x2x^2: the quotient is 12x224+\dfrac{1}{2} - \dfrac{x^2}{24} + \cdots, and every term after the first vanishes as x0x \to 0, leaving 12\dfrac{1}{2}.
  4. 4L'Hopital check: differentiating once gives sinx2x\dfrac{\sin x}{2x}, which is again 0/00/0; differentiating again gives cosx2\dfrac{\cos x}{2}.
  5. 5Substituting x=0x = 0 gives cos02=12\dfrac{\cos 0}{2} = \dfrac{1}{2} — the two methods agree.

limx01cosxx2=12\displaystyle \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}, by series expansion and confirmed by two applications of L'Hopital's rule.

Common traps

  • Applying L'Hopital's rule to a form that is not indeterminate — if the denominator limit is a nonzero number, substitution already answers the question.
  • Stopping L'Hopital's rule after one differentiation when the result is still 0/00/0; the rule may need to be applied again, with the form re-checked each time.
  • Using lim(sinx)/x=1\lim (\sin x)/x = 1 with xx in degrees, or citing L'Hopital to prove that limit when the derivative of sine was derived from it.
  • Concluding a function is continuous because its limit exists — the limit must also equal the function's value at the point.

Question depth and domain coverage vary by exam. Practice answers are checked after submission.

Sources

  1. GRE Subject Test Content and StructureETS. Accessed 2026-07-06. Use as a cited source for exam facts; do not imply affiliation or reproduce protected test material.
  2. Calculus Volume 1, Section 2.3: The Limit LawsOpenStax. Accessed 2026-08-03. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.
  3. Calculus Volume 1, Section 2.4: ContinuityOpenStax. Accessed 2026-08-03. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.
  4. Calculus Volume 1, Section 4.8: L'Hopital's RuleOpenStax. Accessed 2026-08-03. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.
  5. Calculus Volume 2, Section 6.3: Taylor and Maclaurin SeriesOpenStax. Accessed 2026-08-03. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.

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Recheck the ETS content-structure page and the OpenStax Calculus Volume 1 limit and continuity sections before each major GRE Mathematics preparation cycle.