Calculus ILimits and continuity

Continuity, and the Intermediate Value Theorem.

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Learn how to spot a break, join two pieces, and prove a curve reaches a value. Then switch to Solve and work through every class question.

Continuity · Step 1 of 8

Continuous means no hole, no jump, no blow-up

Continuity · Step 1 of 8

Continuous means no hole, no jump, no blow-up

A function is continuous at x = a if you can draw through that point without lifting your pen. Precisely: f(a) is defined, the limit at a exists, and the two are equal, so limx → a f(x) = f(a). This graph from class breaks in five places.

Try it: Pick each x value. Which of the three conditions fail, and what kind of break is it?

Look at x =

1. f(-4) is defined
2. the limit at -4 exists
3. the limit equals f(-4)

Type

Removable

x → −4⁻

x → −4⁺

The limit exists (2), but there is no point at x = −4. Define f(−4) = 2 and the hole is filled.

Removable: a hole

The limit exists but is not f(a), or f(a) is missing. Redefine one point and it is fixed.

Jump: the sides disagree

Both one-sided limits exist but are different.

Infinite: a vertical asymptote

One or both one-sided limits are +∞ or −∞.

Show

Question 1 of 23From class0 of 23 solved

Sine, cosine, and a moved dot

Let

{f(x)} = {\begin{cases}1+\sin x & x<0\\ \cos x & 0\le x\le\pi,\ x\ne\frac{\pi}{2}\\1 & x=\frac{\pi}{2}\\\sin x & x>\pi\end{cases}}

. Find the limits and values, then identify where the two-sided limit fails.

Work it out on paper first, then type your answer. Stuck? Open one step at a time.

(a)

{\lim\limits_{x \to 0^-}f(x)}

Shows here as math.

1

(b)

{\lim\limits_{x \to 0^+}f(x)}

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1

(c)

{f(0)}

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1

(d)

{\lim\limits_{x \to \frac{\pi}{2}^-}f(x)}

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0

(e)

{\lim\limits_{x \to \frac{\pi}{2}^+}f(x)}

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0

(f)

{f\left(\frac{\pi}{2}\right)}

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1

(g)

{\lim\limits_{x \to \pi^-}f(x)}

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-1

(h)

{\lim\limits_{x \to \pi^+}f(x)}

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0

(i)

{f(\pi)}

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-1

(j)

At which input does the two-sided limit not exist?

Shows here as math.

\pi

Stuck on continuity or proving a solution exists?

Bring your practice questions. We work through them together until every type feels routine.

Book a session Next topicNext: Limits at infinity and asymptotes

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Continuity, and the Intermediate Value Theorem.

Continuous means no hole, no jump, no blow-up

Continuous means no hole, no jump, no blow-up

Sine, cosine, and a moved dot

Stuck on continuity or proving a solution exists?

Questions, answered.

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