MediumMathematicsHigh School (11–12)Functions and Calculus › Continuity and DifferentiabilityNumericalBoard Exam

Define f(x) = (1 - cos 4x)/x² for x ≠ 0…

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Define $f(x) = \dfrac{1 - \cos 4x}{x^2}$ for $x \ne 0$ . What value $k = f(0)$ makes $f$ continuous at $x = 0$ ? Use $1 - \cos 4x = 2\sin^2 2x$ and the limit $\dfrac{\sin t}{t} \to 1$ as $t \to 0$ .

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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