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तामिळनाडू बोर्ड ऑफ सेकेंडरी एज्युकेशनएचएससी विज्ञान इयत्ता ११

Evaluate the following limits: elimx→∞×[31x+1-cos(1x)-e1x]

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प्रश्न

Evaluate the following limits:

`lim_(x -> oo) x [3^(1/x) + 1 - cos(1/x) - "e"^(1/x)]`

बेरीज
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उत्तर

We know `lim_(x -> 0) ("e"^x - 1)/x` = 1

`lim_(x -> 0) ("a"^x - 1)/x` = log a

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

`lim_(x -> oo) [3^(1/x) + 1 - cos(1/x) - "e"^(1/x)] =  lim_(x -> oo) [(3^(1/x) + 1 - cos(1/x) - "e"^(1/x))/(1/x)]`

= `lim_(x -> oo) [(3^(1/x) - 1 + 1 - "e"^(1/x))/(1/x) + (1 - cos(1/x))/(1/x)]`

= `lim_(x -> oo) [((3^(1/x) - 1) - ("e"^(1/x) - 1))/(1/x) + (1 - cos(1/x))/(1/x)]`

= `lim_(x > 0)[(3^(1/x) - 1)/(1/x) - ("e"^(1/x) - 1)/(1/x) + (1 -cos(1/x))/(1/x)]`

Put y = `1/x`

When x = `oo`

⇒ y = `1/oo` = 0

`lim_(x -> oo) x [3^(1/x) + 1 - cos(1/x) - "e"^(1/x)] =  lim_(y - 0) [(3y - 1)/y - ("e"^y - 1)/y + (1 - cosy)/y]`

= `(lim_(y -> 0) (3^y - 1)/y) -(lim_(y -> 0) ("e"^y - 1)/y) + (lim_(y -> 0) (1 -  cosy)/y)`

= `log 3 - 1 + 0`

`lim_(x -> oo) x [3^(1/x) + 1 - cos(1/x) - "e"^(1/x)] = (log 3) - 1`

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पाठ 9: Differential Calculus - Limits and Continuity - Exercise 9.4 [पृष्ठ ११८]

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सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
पाठ 9 Differential Calculus - Limits and Continuity
Exercise 9.4 | Q 18 | पृष्ठ ११८

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