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1)... lim (x→0) [ sin ( π cos² x ) ] / x² = lim (x→0) { sin [ π ( 1 - sin² x ) ] } / x² = lim (x→0) [ sin ( π - π sin² x ) ] / x² = lim (x→0) [ sin ( π sin² x ) / x² = lim (x→0) [ ( π sin² x ) / x² ] ....................... from (1) = π • [ lim (x→0) ( (sin x) / x ) ]² = π • [ 1 ]² = π ........................................... Ans.

1)... lim (x→0) [ sin ( π cos² x ) ] / x² 

= lim (x→0) { sin [ π ( 1 - sin² x ) ] } / x² 

= lim (x→0) [ sin ( π - π sin² x ) ] / x² 

= lim (x→0) [ sin ( π sin² x ) / x² 

= lim (x→0) [ ( π sin² x ) / x² ] ....................... from (1) 

= π • [ lim (x→0) ( (sin x) / x ) ]² 

= π • [ 1 ]² 

= π ........................................... Ans. 

Grade:12th pass

1 Answers

Anish Singhal
askIITians Faculty 1194 Points
3 years ago
Yes correct

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