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Lim X → π Sin X π − X

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प्रश्न

\[\lim_{x \to \pi} \frac{\sin x}{\pi - x}\]

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उत्तर

\[\lim_{x \to \pi} \frac{\sin x}{\pi - x}\]
\[ = \lim_{h \to 0} \frac{\sin \left( \pi - h \right)}{\pi - \left( \pi - h \right)} \left[ \because \lim_{x \to a} f\left( x \right) = \lim_{h \to 0} f\left( a - h \right) \right]\]
\[ = \lim_{h \to 0} \frac{\sin h}{h} \left[ \because \sin \left( \pi - 0 \right) = \sin 0 \right]\]
\[ \Rightarrow 1\]

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अध्याय 29: Limits - Exercise 29.8 [पृष्ठ ६२]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 29 Limits
Exercise 29.8 | Q 1 | पृष्ठ ६२

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