Evaluate the Given limit: \( \lim\limits_{x \to π} \dfrac{sin( π-x)}{ π( π-x)} \)
\( \lim\limits_{x \to π} \dfrac{sin( π-x)}{ π( π-x)} \)
It is seen that x → π ⇒ (π - x) → 0
= \(\dfrac{1}{π} \lim\limits_{π-x \to 0} \dfrac{sin( π-x)}{( π-x)} \)
= \( \dfrac{1}{π}\times 1\)
= \( \dfrac{1}{π}\)
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