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तमिलनाडु बोर्ड ऑफ सेकेंडरी एज्युकेशनएचएससी विज्ञान कक्षा ११

Evaluate the following limits: limx→πsin3xsin2x

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प्रश्न

Evaluate the following limits:

`lim_(x -> pi) (sin3x)/(sin2x)`

योग
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उत्तर

`lim_(x -> pi) (sin3x)/(sin2x) =  lim_(x -> pi) (3sin x - 4 sin^3 x)/(2sinx cos  x)`

= `lim_(x -> pi) [(3sinx)/(2sinx cosx) - (4sin^3x)/(sinx cosx)]`

= `lim_(x -> pi) [3/(2cosx) - (2sin^2x)/cosx]`

= `lim_(x -> pi) 3/(2cosx) - lim_(x -> pi) (2sin^2x)/cosx`

= `3/(2cospi) - (2sin^2pi)/cospi`

=`3/(2 xx -1) - (2 xx 0)/(-1)`

`lim_(x -> pi) (sin3x)/(sin2x) = - 3/2`

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 9: Differential Calculus - Limits and Continuity - Exercise 9.4 [पृष्ठ ११८]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
अध्याय 9 Differential Calculus - Limits and Continuity
Exercise 9.4 | Q 20 | पृष्ठ ११८

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