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Checking continuity at a point using the three-part definition

Consider the function n(x)={1−cos⁡xx2x≠012x=0n(x) = \begin{cases} \dfrac{1-\cos x}{x^2} & x \neq 0 \\ \dfrac{1}{2} & x = 0 \end{cases}

Use the three-part definition of continuity to determine whether n(x)n(x) is continuous at x=0x = 0. In your answer, state the value of n(0)n(0), the value of lim⁡x→0n(x)\lim_{x \to 0} n(x), and whether these two conditions plus the existence of the limit are all satisfied. Conclude whether nn is continuous at x=0x=0.

Hint

Recall the three checks: (1) is n(0)n(0) defined? (2) does lim⁡x→0n(x)\lim_{x\to 0} n(x) exist? (3) does that limit equal n(0)n(0)? You may want to recall or derive that lim⁡x→01−cos⁡xx2=12\lim_{x\to 0} \frac{1-\cos x}{x^2} = \frac{1}{2}, for instance by using 1−cos⁡x=2sin⁡2(x/2)1-\cos x = 2\sin^2(x/2) and the known limit lim⁡θ→0sin⁡θθ=1\lim_{\theta\to 0} \frac{\sin\theta}{\theta}=1.

Step-by-step solution

First, check that n(0)n(0) is defined: by the piecewise definition, n(0)=12n(0) = \frac{1}{2}, so condition 1 is satisfied. Next, evaluate lim⁡x→0n(x)\lim_{x\to 0} n(x) using the branch 1−cos⁡xx2\frac{1-\cos x}{x^2} for x≠0x \neq 0. Using the identity 1−cos⁡x=2sin⁡2(x/2)1-\cos x = 2\sin^2(x/2), we get 1−cos⁡xx2=2sin⁡2(x/2)x2=12(sin⁡(x/2)x/2)2\frac{1-\cos x}{x^2} = \frac{2\sin^2(x/2)}{x^2} = \frac{1}{2}\left(\frac{\sin(x/2)}{x/2}\right)^2.

As x→0x \to 0, sin⁡(x/2)x/2→1\frac{\sin(x/2)}{x/2} \to 1, so the whole expression tends to 12(1)2=12\frac{1}{2}(1)^2 = \frac{1}{2}. Thus lim⁡x→0n(x)=12\lim_{x\to 0} n(x) = \frac{1}{2} exists, satisfying condition 2. Since this limit equals n(0)=12n(0) = \frac{1}{2}, condition 3 is also satisfied, so all three conditions hold and nn is continuous at x=0x=0.

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