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Exercise 12.1 · Q15

Q.lim⁡x→πsin⁡(π−x)π(π−x)\lim_{x\to \pi}\dfrac{\sin(\pi - x)}{\pi(\pi - x)}

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Substitute t=π−xt=\pi-x so that t→0t\to 0; the expression becomes 1π⋅sin⁡tt\dfrac{1}{\pi}\cdot\dfrac{\sin t}{t}, and using the standard limit sin⁡tt→1\dfrac{\sin t}{t}\to 1 gives 1π\dfrac{1}{\pi}.

Why a substitution turns this into a standard limit

The numerator sin⁡(π−x)\sin(\pi-x) and the denominator both contain the same quantity (π−x)(\pi-x), which tends to 00 as x→πx\to\pi. Whenever we see sin⁡(something)\sin(\text{something}) over that same something\text{something}, the standard limit lim⁡t→0sin⁡tt=1\lim_{t\to 0}\dfrac{\sin t}{t}=1 is the right tool.

Step 1 — Substitute. Let t=π−xt=\pi-x. As x→πx\to\pi, t→0t\to 0, and

sin⁡(π−x)π(π−x)=sin⁡tπ t.\dfrac{\sin(\pi-x)}{\pi(\pi-x)}=\dfrac{\sin t}{\pi\, t}.

Step 2 — Pull out the constant 1π\dfrac{1}{\pi}.

lim⁡t→0sin⁡tπ t=1πlim⁡t→0sin⁡tt.\lim_{t\to 0}\dfrac{\sin t}{\pi\,t}=\dfrac{1}{\pi}\lim_{t\to 0}\dfrac{\sin t}{t}. …

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