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Evaluate the following : limx→0[x(6x-3x)cos(6x)-cos(4x)] - Mathematics and Statistics

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प्रश्न

Evaluate the following :

`lim_(x -> 0) [(x(6^x - 3^x))/(cos (6x) - cos (4x))]`

योग
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उत्तर

`lim_(x -> 0) (x(6^x - 3^x))/(cos (6x) - cos (4x))`

= `lim_(x -> 0) (x*3^x (2^x - 1))/(-2sin ((6x + 4x)/2) sin((6x - 4x)/2))`

= `1/(-2) lim_(x -> 0) (x*3^x (2^x - 1))/(sin 5x*sinx)`

= `1/(-2) lim_(x -> 0) ((x*3^x (2^x - 1))/x^2)/((sin 5x* sinx)/x^2)   ...[("Divide Numerator and"),("Denominator by"  x^2),(∵ x -> 0"," ∴ x ≠ 0 ∴ x^2 ≠ 0)]`

= `1/(-2) (lim_(x -> 0) [3^x ((2^x - 1))/x])/(lim_(x -> 0) ((sin5x)/x * sinx/x))`

= `1/(-2) (lim_(x -> 0)(3x) * lim_(x -> 0) ((2^x - 1)/x))/(lim_(x -> 0) (sinx/(5x) xx 5) * lim_(x -> 0) (sinx/x))`

= `1/(-2) xx (3^circ xx log 2)/(1 xx 5 xx 1)  ...[(because x -> 0","  5x -> 0),(lim_(x -> 0) ("a"^x - 1)/x = log "a"),(lim_(x -> 0) sinx/x = 1)]`

= `(-1)/10 log 2`

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 7: Limits - Miscellaneous Exercise 7.2 [पृष्ठ १५९]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 11 Maharashtra State Board
अध्याय 7 Limits
Miscellaneous Exercise 7.2 | Q II. (9) | पृष्ठ १५९

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