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महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता ११ वी

Evaluate the following : limx→a[xcosa-acosxx-a]

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प्रश्न

Evaluate the following :

`lim_(x -> "a") [(x cos "a" - "a" cos x)/(x - "a")]`

बेरीज
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उत्तर

`lim_(x -> "a") [(x cos "a" - "a" cos x)/(x - "a")]`

= `lim_(x -> "a") [(x cos "a" - "a" cos "a" + "a" cos "a" - "a" cos x)/(x - "a")]`   ...[Note this step]

= `lim_(x -> "a") [((x - "a") cos "a" + "a"(cos"a" - cosx))/(x - "a")]`

= `lim_(x -> "a") [((x - "a") cos "a" + 2"a" sin (("a" + x)/2)((x - "a")/2))/(x - "a")]`

= `lim_(x -> "a") [((x - "a")cos"a")/(x - "a") + (2"a" sin (("a" + x)/2) sin((x - "a")/2))/(x - "a")]`

= `lim_(x -> "a") [cos"a" + "a" sin (("a" + x)/2)* (sin((x - "a")/2))/(((x - "a")/2))]`  ...[∵ x → a, x ≠ a, ∴ x – a ≠ 0]

= `lim_(x -> "a") cos"a" + "a"[lim_(x -> "a") sin(("a" + x)/2)] xx [lim_(x -> "a") sin((x - "a")/2)/((x - "a")/2)]` 

= `cos "a" + "a" sin (("a" + "a")/2) xx 1     ...[(because x -> "a" ","  x ≠ "a"  therefore (x - "a")/2 -> 0),(and lim_(theta -> 0) sintheta/theta = 1)]`

= cos a + a sin a

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  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 7: Limits - Miscellaneous Exercise 7.2 [पृष्ठ १५९]

APPEARS IN

बालभारती Mathematics and Statistics (Arts and Science) Part 2 [English] Standard 11 Maharashtra State Board
पाठ 7 Limits
Miscellaneous Exercise 7.2 | Q II. (16) | पृष्ठ १५९

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