How does f(x) = x^2 if x ≠ 2; 5 if x = 2 behave at x = 2?

Continuity & Differentiability · Question 13 of 125

Question 13

MediumOriginal question

How does $f(x) = \begin{cases} x^{2}, & x \ne 2 \\ 5, & x = 2 \end{cases}$ behave at $x = 2$ ?

Continuous at the pointCorrect answerYour answerNot continuous: the left and right limits differCorrect answerYour answerNot continuous: the function has no value thereCorrect answerYour answerNot continuous: the limit exists but differs from the valueCorrect answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer D

Correct answer: D

Continuity at $a$ needs the left limit, the right limit and $f(a)$ to be equal.
Here the left limit is $4$ , the right limit $4$ and $f(2) = 5$ .

Worth noting

The limit 4 exists, but the function is given the value 5 there.

Concept: Testing continuity at a point
Typical timeTypical: 90 s
Your timeYours: —

The rule this testsContinuity at a point

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How does f(x)={x2,x≠25,x=2f(x) = \begin{cases} x^{2}, & x \ne 2 \\ 5, & x = 2 \end{cases}f(x)={x2,5,​x=2x=2​ behave at x=2x = 2x=2?

Explanation

How does $f(x) = \begin{cases} x^{2}, & x \ne 2 \\ 5, & x = 2 \end{cases}$ behave at $x = 2$ ?