Continuity & Uniform Convergence Practice Quiz with a Step-by-Step Interactive Lesson

Continuity at a point: \(\varepsilon\)–\(\delta\), limits, and common pitfalls

Key idea

A function \(f\) is continuous at \(a\) if: \[ \forall \varepsilon > 0 \ \exists \delta > 0 \ \text{such that}\ |x-a|<\delta \Rightarrow |f(x)-f(a)|<\varepsilon. \] If \(\lim_{x\to a} f(x)\) exists, then continuity at \(a\) is equivalent to: \[ \lim_{x\to a} f(x)=f(a). \] To check continuity quickly, you often:

• Compute the limit \(\lim_{x\to a} f(x)\) (two-sided or one-sided if \(a\) is an endpoint).
• Compare it to the defined value \(f(a)\).
• Identify whether any “break” is removable, a jump, or an infinite blow-up.

Worked example

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Summary

• Continuity at \(a\): \(\lim_{x\to a} f(x)=f(a)\) (when the limit exists).
• Many discontinuities are detected by comparing one-sided limits and the function value.

Continuity on \([a,b]\), one-sided limits, and max/min theorems

Key idea

To say \(f\) is continuous on \([a,b]\) means:

• \(f\) is continuous at every interior point \(x\in(a,b)\), and
• \(f\) is right-continuous at \(a\) and left-continuous at \(b\) (one-sided continuity at endpoints).

Two foundational results for continuous functions on closed intervals:

• Extreme Value Theorem (EVT): If \(f\) is continuous on \([a,b]\), then \(f\) attains a maximum and a minimum on \([a,b]\).
• Intermediate Value Theorem (IVT): If \(f\) is continuous on \([a,b]\), then \(f\) takes every value between \(f(a)\) and \(f(b)\).

Worked example

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Summary

• EVT and IVT are guaranteed on closed intervals \([a,b]\), not on open intervals \((a,b)\).
• Continuity is preserved under composition and under taking absolute value.

Uniform continuity vs. continuity: what changes and why it matters

Key idea

Continuity at a point \(a\) allows \(\delta\) to depend on \(a\). Uniform continuity strengthens this by requiring one \(\delta\) that works for the whole set: \[ \forall \varepsilon>0\ \exists \delta>0\ \text{such that for all }x,y\in E,\ |x-y|<\delta \Rightarrow |f(x)-f(y)|<\varepsilon. \] That “one \(\delta\) for all points” is the key phrase for uniform continuity.

Two must-know facts

• Heine–Cantor theorem: If \(f\) is continuous on a compact interval \([a,b]\), then \(f\) is uniformly continuous on \([a,b]\).
• On unbounded sets: being continuous does not guarantee uniform continuity (e.g., \(x^2\) on \(\mathbb{R}\)).

Worked example

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Summary

• Uniform continuity is “continuity with \(\delta\) independent of the point.”
• Continuous on \([a,b]\) \(\Rightarrow\) uniformly continuous on \([a,b]\) (Heine–Cantor).

Pointwise convergence vs. uniform convergence: the sup norm test

Key idea

Pointwise convergence means: for each fixed \(x\in E\), \(f_n(x)\to f(x)\). Uniform convergence means: \[ \sup_{x\in E}\,|f_n(x)-f(x)| \to 0. \] Equivalently, \(f_n\to f\) uniformly on \(E\) if: \[ \forall \varepsilon>0\ \exists N\ \text{such that } n\ge N \Rightarrow |f_n(x)-f(x)|<\varepsilon \text{ for all }x\in E. \] The key difference: in uniform convergence, the \(N\) works for all \(x\) simultaneously.

Worked example

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Summary

• Uniform convergence uses \(\sup_{x\in E}|f_n-f|\to 0\).
• \(x^n\to 0\) uniformly on \([0,c]\) for \(c<1\), but not uniformly on \([0,1]\).

Uniform convergence of \(\sum f_n\): Weierstrass M-test and fast bounds

Key idea

A series of functions \(\sum_{n=1}^{\infty} f_n(x)\) is said to converge uniformly on a set \(E\) if the sequence of partial sums \[ S_N(x)=\sum_{n=1}^{N} f_n(x) \] converges uniformly to a limit function \(S(x)\) on \(E\). The most common test is the Weierstrass M-test:

• If \(|f_n(x)| \le M_n\) for all \(x\in E\), and \(\sum_{n=1}^{\infty} M_n\) converges, then \(\sum f_n\) converges uniformly (and absolutely) on \(E\).

Worked example

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Summary

• M-test is a go-to tool: bound \(|f_n(x)|\) by a convergent numerical series.
• Uniform limits behave well with respect to sup-norm bounds.

Does the limit stay continuous? Uniform convergence vs. pointwise convergence

Key idea

Uniform convergence preserves continuity: If each \(f_n\) is continuous on \(E\) and \(f_n\to f\) uniformly on \(E\), then \(f\) is continuous on \(E\). This is one of the main reasons uniform convergence is stronger than pointwise convergence.

Classic counterexample (pointwise only)

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Summary

• Uniform convergence preserves continuity; pointwise convergence does not.
• \(x^n\) on \([0,1]\) is a key example: continuous \(f_n\), discontinuous pointwise limit.

How to solve continuity & uniform convergence questions fast

Fast checklist (what to try first)

• Continuity at \(a\): compute \(\lim_{x\to a} f(x)\) and compare to \(f(a)\). For piecewise functions, check one-sided limits.
• Continuity on an interval: verify denominator never hits 0; ensure endpoints use one-sided continuity on \([a,b]\).
• Uniform continuity: on \([a,b]\), use Heine–Cantor. On \(\mathbb{R}\), be cautious with unbounded slope.
• Uniform convergence: compute \(\sup |f_n-f|\) or use a bounding argument. For \(\sum f_n\), try the M-test.
• Preservation facts: uniform convergence preserves continuity; pointwise convergence does not.

Worked example: fixing a removable discontinuity

Final check

Final recap

• Continuity at \(a\): \(\forall \varepsilon>0\ \exists \delta>0\) so \(|x-a|<\delta \Rightarrow |f(x)-f(a)|<\varepsilon\).
• Uniform continuity: \(\forall \varepsilon>0\ \exists \delta>0\) so \(|x-y|<\delta \Rightarrow |f(x)-f(y)|<\varepsilon\) for all \(x,y\in E\).
• Uniform convergence: \(\sup_{x\in E}|f_n(x)-f(x)|\to 0\).
• Uniform convergence + continuity of \(f_n\) \(\Rightarrow\) continuity of \(f\).
• M-test: if \(|f_n(x)|\le M_n\) and \(\sum M_n\) converges, then \(\sum f_n\) converges uniformly.