Which one of the following functions is not c | vedclass

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(D) We analyze the continuity of each function on the interval $(0, \pi )$: $1$. $f(x) = \cot x = \frac{\cos x}{\sin x}$. This function is discontinuo

Vedclass · Accepted Answer

$l(x) = \begin{cases} x \sin x & 0 < x \le \frac{\pi}{2} \ \frac{\pi}{2} \sin(x + \pi) & \frac{\pi}{2} < x < \pi \end{cases}$

Vedclass · Answer

(D) We analyze the continuity of each function on the interval $(0, \pi )$: $1$. $f(x) = \cot x = \frac{\cos x}{\sin x}$. This function is discontinuous at $x = \pi /2$ is incorrect,it is discontinuous at $x = n\pi$. Within $(0, \pi )$,$\sin x$ is never $0$,so $f(x)$ is continuous. $2$. $g(x) = \int_{0}^{x} t \sin \frac{1}{t} \, dt$. Since the integrand $t \sin(1/t)$ is bounded and continuous on $(0, \pi )$,the integral function $g(x)$ is continuous on $(0, \pi )$. $3$. $h(x)$ is defined piecewise. At $x = 3\pi /4$,the left-hand limit is $1$. The right-hand limit is $2 \sin(\frac{2}{9} \cdot \frac{3\pi}{4}) = 2 \sin(\frac{\pi}{6}) = 2(1/2) = 1$. Since the limits match,$h(x)$ is continuous. $4$. $l(x)$ is defined piecewise. At $x = \pi /2$,the left-hand limit is $\frac{\pi}{2} \sin(\frac{\pi}{2}) = \frac{\pi}{2}$. The right-hand limit is $\frac{\pi}{2} \sin(\frac{\pi}{2} + \pi) = \frac{\pi}{2} \sin(\frac{3\pi}{2}) = \frac{\pi}{2}(-1) = -\frac{\pi}{2}$. Since $\frac{\pi}{2} 
eq -\frac{\pi}{2}$,$l(x)$ is discontinuous at $x = \pi /2$.

Which one of the following functions is not continuous on $(0, \pi )$?

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