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तमिलनाडु बोर्ड ऑफ सेकेंडरी एज्युकेशनएचएससी विज्ञान कक्षा ११

Evaluate the following limits: limx→0tanx-sinxx3

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प्रश्न

Evaluate the following limits:

`lim_(x -> 0) (tan x - sin x)/x^3`

योग
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उत्तर

We know `lim_(x -> 0) sinx/x` = 1

`lim_(x -> 0) (tan x - sin x)/x^3 =  lim_(x -> 0) (sinx/cosx - sin x)/x^3`

= `lim_(x -> 0) ((sinx - sinx cosx)/cosx)/x^3`

= `lim_(x -> 0) (sinx(1 -  cosx))/(x^3 cosx)`

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

= `lim_(x -> 0) sinx/x xx (2sin^2 (x/2))/(2^2 xx x^2/2^2) xx 1/cosx`

= `lim_(x -> 0) sinx/x xx 1/2 (lim_(x/2 -> 0) (sin(x/2))/(x/2))^2 xx lim_(x - 0) 1/cosx`

= `1 xx 1/2 xx 1^2 xx 1/cos0`

= `1/2 xx 1/1`

`lim_(x -> 0) (tan x - sin x)/x^3 = 1/2`

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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 28 | पृष्ठ ११८

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