Determine whether t(x) is continuous at x = π/2. Use the three-part definition of continuity: (1) is t(π/2) defined, (2) does limx → π/2 t(x) exist, and (3) does the limit equal t(π/2)? Based on your work, state whether t is continuous at x = π/2, and if it is not, identify which condition fails and classify the type of discontinuity (removable, jump, or infinite).

First check condition (1): since x = π/2 falls under the rule 1-x (because that piece applies for x ≥ π/2), we have t(π/2) = 1 - π/2, so t(π/2) is defined. Next check condition (2): compute the one-sided limits. The left-hand limit uses cos(x) since that piece applies for $x

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Checking the three-part continuity test at a piecewise boundary

Consider the piecewise function $t(x) = \begin{cases} \cos(x) & x < \pi/2 \\ 1 - x & x \ge \pi/2 \end{cases}$

Determine whether $t(x)$ is continuous at $x = \pi/2$ . Use the three-part definition of continuity: (1) is $t(\pi/2)$ defined, (2) does $\lim_{x \to \pi/2} t(x)$ exist, and (3) does the limit equal $t(\pi/2)$ ? Based on your work, state whether $t$ is continuous at $x = \pi/2$ , and if it is not, identify which condition fails and classify the type of discontinuity (removable, jump, or infinite).

Hint

Since $t$ is defined by different formulas on either side of $\pi/2$ , compute the left-hand limit using $\cos(x)$ and the right-hand limit using $1-x$ separately, then compare both to the value $t(\pi/2)$ given by the piece that includes $\pi/2$ .

Step-by-step solution

First check condition (1): since $x = \pi/2$ falls under the rule $1-x$ (because that piece applies for $x \ge \pi/2$ ), we have $t(\pi/2) = 1 - \pi/2$ , so $t(\pi/2)$ is defined. Next check condition (2): compute the one-sided limits. The left-hand limit uses $\cos(x)$ since that piece applies for $x < \pi/2$ , giving $\lim_{x \to (\pi/2)^-} \cos(x) = \cos(\pi/2) = 0$ . The right-hand limit uses $1-x$ , giving $\lim_{x \to (\pi/2)^+} (1-x) = 1 - \pi/2 \approx -0.5708$ .

Since the left-hand limit $(0)$ and right-hand limit $(1-\pi/2 \approx -0.5708)$ are not equal, $\lim_{x\to \pi/2} t(x)$ does not exist, so condition (2) fails. Because both one-sided limits exist (they're just unequal, and finite), this is a jump discontinuity — the graph literally jumps from height $0$ to height about $-0.5708$ at $x=\pi/2$ . Therefore $t(x)$ is not continuous at $x = \pi/2$ .

Checking the three-part continuity test at a piecewise boundary — solution graph

Checking the three-part continuity test at a piecewise boundary

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