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Tamil Nadu Board of Secondary EducationHSC Science Class 11

Evaluate the following limits: limx→∞(2x2+32x2+5)8x2+3

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Question

Evaluate the following limits:

`lim_(x -> oo) ((2x^2 + 3)/(2x^2 + 5))^(8x^2 + 3)`

Sum
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Solution

`lim_(x -> oo) ((2x^2 + 3)/(2x^2 + 5))^(8x^2 + 3) =  lim_(x -> oo)((2x^2 + 5 - 2)/(2x^2 + 5))^(8x^2 + 20 - 17)`

= `lim_(x -> oo) ((2x^2 - 5)/(2x^2 + 5) - 2/(2x^2 + 5))^(4(2x^2 + 5) - 17)`

= `lim_(x -> 00) (1 - 2/(2x^2 + 5))^(4(2x^2 + 5) - 17)`

Put 2x2 + 5 = y

When x → ∞

We have y = 2 × ∞ + 5 = ∞

x → ∞

⇒ y → ∞

∴ `lim_(x -> oo) ((2x^2 + 3)/(2x^2 + 5))^(8x^2 + 3) =  lim_(y -> oo) (1 - 2/y)^(4y - 17)`

= `lim_(y - oo) (1 - 2/y)^(4y) xx (1 - 2/y)^(-17)`

= `lim_(y ->oo) (1 -2/y)^(4y) xx lim_(y -> oo) (1 - 2/y)^(- 17)`

= `(lim_(y -> oo) (1 - 2/y)^y)^4 xx (1 - 2/oo)^(- 17)`  ........(1)

We know `lim_(x -> oo) (1 + "k/x)^x` = ek

(1) ⇒ `lim_(x -> oo) ((2x^2 + 3)/(2x^2 + 5))^(8x^2 + 3)`

= `(lim_(y -> oo)(1 + ((-2))/y)^y)^4 xx (1 - 0)^(- 17)`

= `("e"^(-2))^4 xx 1`

= `"e"^(-8)`

= `1/"e"^8`

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Chapter 9: Differential Calculus - Limits and Continuity - Exercise 9.4 [Page 117]

APPEARS IN

Samacheer Kalvi Mathematics - Volume 1 and 2 [English] Class 11 TN Board
Chapter 9 Differential Calculus - Limits and Continuity
Exercise 9.4 | Q 4 | Page 117

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