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`lim_(x->pi)(sin(pi-x)/(pi(pi-x)))` |
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Answer» Correct Answer - `(1)/(pi)` Putting `(pi-x)=h, "we have" x to pi implies pi-xto0rarrhto0.` `thereforelim_(xto pi)(sin(pi-x))/(pi(pi-x))=(1)/(pi)lim_(hto0)(sinh)/(h)=((1)/(pi)xx1)=(1)/(pi).` |
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