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Lim X → 1 Sin π X X − 1

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प्रश्न

\[\lim_{x \to 1} \frac{\sin \pi x}{x - 1}\] 

विकल्प

  • π 

  • π 

  • \[- \frac{1}{\pi}\] 

  • \[\frac{1}{\pi}\] 

MCQ
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उत्तर

 −π 

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

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 29: Limits - Exercise 29.13 [पृष्ठ ७९]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 29 Limits
Exercise 29.13 | Q 17 | पृष्ठ ७९

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