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h(x)={[cos(x)," for "x < pi],[sin(x)," for "x >= pi]:} Find lim_(x rarrpi^(+))h(x). Choose 1 answer: (A) -1 (B) 0 (c) 1 (D) The limit doesn't exist.

h(x)={cos⁡(x)amp; for xlt;πsin⁡(x)amp; for x≥π h(x)=\left\{\begin{array}{ll} \cos (x) &amp; \text { for } x&lt;\pi \\ \sin (x) &amp; \text { for } x \geq \pi \end{array}\right. \newlineFind lim⁡x→π+h(x) \lim _{x \rightarrow \pi^{+}} h(x) .\newlineChoose 11 answer:\newline(A) −1-1\newline(B) 00\newline(C) 11\newline(D) The limit doesn't exist.

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Full solution

Q. h(x)={cos⁡(x) for x<πsin⁡(x) for x≥π h(x)=\left\{\begin{array}{ll} \cos (x) & \text { for } x<\pi \\ \sin (x) & \text { for } x \geq \pi \end{array}\right. \newlineFind lim⁡x→π+h(x) \lim _{x \rightarrow \pi^{+}} h(x) .\newlineChoose 11 answer:\newline(A) −1-1\newline(B) 00\newline(C) 11\newline(D) The limit doesn't exist.
  1. Problem Statement: We are asked to find the limit of the function h(x)h(x) as xx approaches π\pi from the right, which is denoted as lim⁡x→π+h(x)\lim_{x \to \pi^+} h(x). To do this, we need to look at the definition of the function h(x)h(x) for values of xx that are greater than or equal to π\pi.
  2. Definition of h(x): According to the definition of h(x), for x≥πx \geq \pi, h(x)=sin⁡(x)h(x) = \sin(x). Therefore, to find the limit as xx approaches π\pi from the right, we need to evaluate the sine function at π\pi.
  3. Evaluation of h(x)h(x) at π\pi: The sine of π\pi is 00. Therefore, lim⁡x→π+h(x)=sin⁡(π)=0\lim_{x \to \pi^+} h(x) = \sin(\pi) = 0.

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