How does f(x) = (x^3 - 8)/(x - 2) if x ≠ 2; 12 if x = 2 behave at x = 2?

Question 25

HardOriginal question

How does $f(x) = \begin{cases} \frac{x^{3} - 8}{x - 2}, & x \ne 2 \\ 12, & x = 2 \end{cases}$ behave at $x = 2$ ?

Not continuous: the limit exists but differs from the valueCorrect answerYour answerContinuous at the pointCorrect answerYour answerNot continuous: the function has no value thereCorrect answerYour answerNot continuous: the left and right limits differCorrect answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer B

Correct answer: B

Continuity at $a$ needs the left limit, the right limit and $f(a)$ to be equal.
Here the left limit is $12$ , the right limit $12$ and $f(2) = 12$ .

Worth noting

At $x = 2$ the formula is $\frac{0}{0}$ , but the limit is $x^{2} + 2x + 4 = 12$ , the value given.

Concept: Testing continuity at a point
Typical timeTypical: 120 s
Your timeYours: —

The rule this testsContinuity at a point

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How does f(x)={x3−8x−2,x≠212,x=2f(x) = \begin{cases} \frac{x^{3} - 8}{x - 2}, & x \ne 2 \\ 12, & x = 2 \end{cases}f(x)={x−2x3−8​,12,​x=2x=2​ behave at x=2x = 2x=2?

Explanation

How does $f(x) = \begin{cases} \frac{x^{3} - 8}{x - 2}, & x \ne 2 \\ 12, & x = 2 \end{cases}$ behave at $x = 2$ ?