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A function is defined by f(x) = (x³ - 8)/(x - 2) for x ≠ 2, and f(2) = 5…

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A function is defined by $f(x) = \dfrac{x^3 - 8}{x - 2}$ for $x \ne 2$ , and $f(2) = 5$ . Find the limit of $f(x)$ as $x \to 2$ . Then find how far the value $f(2)$ is from the limit.

Need a reminder? The key ideaShow

A function $f$ is continuous at $a$ when $\displaystyle\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a)$ . If the one-sided limits differ, the break is a jump. If the limit exists but $f(a)$ differs, the break is removable.

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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