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Exercise 9.1 · Q17

Q.Sketch the graph of ff, then identify the values of x0x_0 for which lim⁡x→x0f(x)\displaystyle\lim_{x\to x_0}f(x) exists.
[!FORMULA] f(x)={sin⁡x,x<01−cos⁡x,0≤x≤πcos⁡x,x>πf(x)=\begin{cases}\sin x, & x<0\\ 1-\cos x, & 0\le x\le\pi\\ \cos x, & x>\pi\end{cases}

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Step 1. Identify the pieces and their shapes. f(x)=sin⁡xf(x)=\sin x for x<0x<0; f(x)=1−cos⁡xf(x)=1-\cos x for 0≤x≤π0\le x\le\pi (a "hump" from 00 up to 22 at x=πx=\pi); f(x)=cos⁡xf(x)=\cos x for x>πx>\pi. To sketch: draw the sine curve for negative xx, the 1−cos⁡x1-\cos x hump on [0,π][0,\pi], and the cosine curve for x>πx>\pi.

Step 2. Check the first junction, x=0x=0. From the left (x<0x<0, using sin⁡x\sin x): as x→0−x\to0^-, f(x)→sin⁡0=0f(x)\to\sin0=0. From the right (0≤x≤π0\le x\le\pi, using 1−cos⁡x1-\cos x): as x→0+x\to0^+, f(x)→1−cos⁡0=1−1=0f(x)\to1-\cos0=1-1=0. Both sides agree at 00, so lim⁡x→0f(x)=0\displaystyle\lim_{x\to0}f(x)=0 exists. …

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