Continuity and Discontinuity Practice

Sample questions

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1. Sample 1difficulty 1/5

   A function is continuous at $x = a$ on a graph when

   • A

     The function is differentiable at $x = a$

   • B

     The graph has no break at $x = a$

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   • C

     The function has a horizontal tangent at $x = a$

   • D

     The function attains its maximum at $x = a$

   Why

   Continuity means the graph can be drawn through the point without lifting the pencil — no jumps, holes, or asymptotes.

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2. Sample 2difficulty 2/5

   The graph has $\lim_{x\to 1^-} f = 2$, $\lim_{x\to 1^+} f = 4$, $f(1) = 4$. Which is true?

   • A

     $\lim_{x\to 1} f$ does not exist (jump)

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   • B

     $\lim_{x\to 1} f$ exists

   • C

     $f$ is undefined at $1$

   • D

     $f$ is continuous at $1$

   Why

   Left and right one-sided limits differ → two-sided limit DNE.

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3. Sample 3difficulty 2/5

   $g(x) = \begin{cases} 2x & x < 0 \\ x^2 + c & x \ge 0 \end{cases}$. For continuity at $x = 0$, $c$ must be

   • A

     $1$

   • B

     $2$

   • C

     $0$

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   • D

     $-2$

   Why

   Left: $\lim_{x\to 0^-} 2x = 0$. Right: $0^2 + c = c$. So $c = 0$.

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4. Sample 4difficulty 2/5

   For $f$ to be continuous at $x = a$, which must hold?

   • A

     $\lim_{x\to a} f(x)$ exists

   • B

     $f(a)$ is defined

   • C

     All three of the above

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   • D

     $\lim_{x\to a} f(x) = f(a)$

   Why

   All three conditions of the definition of continuity at a point.

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5. Sample 5difficulty 3/5

   Let $f(x) = \begin{cases} \dfrac{x^2 - 4}{x - 2} & x \ne 2 \\ k & x = 2 \end{cases}$. Find $k$ so that $f$ is continuous at $x = 2$.

   • A

     $2$

   • B

     $0$

   • C

     Cannot be made continuous

   • D

     $4$

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   Why

   For $x \ne 2$, the function reduces to $x + 2$, with limit $4$ as $x \to 2$. So set $k = 4$.

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