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Checking the three-part definition of continuity at a point

Consider the function e(x)={tan⁡(x)x≠π/41x=π/4e(x) = \begin{cases} \tan(x) & x \neq \pi/4 \\ 1 & x = \pi/4 \end{cases}

Use the three-part definition of continuity to determine whether e(x)e(x) is continuous at x=π/4x = \pi/4. In your answer, state the value of e(π/4)e(\pi/4), the value of lim⁡x→π/4e(x)\lim_{x \to \pi/4} e(x), and whether these two values are equal. Based on this, is e(x)e(x) continuous at x=π/4x = \pi/4? If it is discontinuous, classify the discontinuity as removable, jump, or infinite.

Hint

Check each of the three conditions separately: is e(π/4)e(\pi/4) defined, does the limit as x→π/4x \to \pi/4 exist, and if so, does it match e(π/4)e(\pi/4)? Recall that away from x=π/4x=\pi/4, e(x)e(x) is just tan⁡(x)\tan(x), which is continuous there.

Step-by-step solution

First, evaluate the function at the point: by definition, e(π/4)=1e(\pi/4) = 1, so condition 1 is satisfied. Next, find the limit as x→π/4x \to \pi/4. Since e(x)=tan⁡(x)e(x) = \tan(x) for all xx near π/4\pi/4 except at the point itself, and tan⁡(x)\tan(x) is continuous at x=π/4x=\pi/4, we have lim⁡x→π/4e(x)=lim⁡x→π/4tan⁡(x)=tan⁡(π/4)=1\lim_{x \to \pi/4} e(x) = \lim_{x \to \pi/4} \tan(x) = \tan(\pi/4) = 1. So the limit exists and equals 11, satisfying condition 2.

Now compare the limit to the function value: lim⁡x→π/4e(x)=1\lim_{x \to \pi/4} e(x) = 1 but e(π/4)=1e(\pi/4) = 1 as well... wait, checking definitions carefully, e(π/4)e(\pi/4) is defined to be 11, which happens to match the limit value tan⁡(π/4)=1\tan(\pi/4)=1. Since all three conditions hold (e(π/4)e(\pi/4) defined, limit exists, and limit equals e(π/4)e(\pi/4)), the function e(x)e(x) is continuous at x=π/4x = \pi/4.

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