`lim_(x->oo) ((2x - 3)/(2x + 5))^(2x + 1)` Find the limit. Use l’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it. If l’Hospital’s Rule...

`lim_(x->oo) ((2x - 3)/(2x + 5))^(2x + 1)` Find the limit. Use l’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it. If l’Hospital’s Rule doesn’t apply, explain why.

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Textbook Question

Chapter 4, 4.4 - Problem 66 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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sciencesolve | Teacher | (Level 3) Educator Emeritus

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Replacing `oo` for x in limit equation yields the nedetermination `oo^oo` . You need to use the following technique, such that:

`f(x) = ((2x - 3)/(2x + 5))^(2x + 1)`

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

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

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

Since `lim_(x->oo) (- 8)/(2x + 5) = -8/oo = 0,` yields:

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

You need to remember that `lim_(x->oo) (1 + 1/x)^x = e` , hence, you need to re-create the special limit:

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

You need to notice that `lim_(x->oo)(1+1/x)^x = oo` , such that:

` lim_(x->oo)((1 + (-8)/(2x+5))^(((2x+5)/(-8))))^(-8(2x+1)/(2x+5)) = e^lim_(x->oo) (-8(2x+1)/(2x+5)) `

You need to evaluate `lim_(x->oo)(- 8)*(2x+1)/(2x + 5)` :

`lim_(x->oo)(- 8)*(2x+1)/(2x + 5) = -8*lim_(x->oo)(2x+1)/(2x + 5) = -8*(2/2) = -8`

`e^lim_(x->oo)(- 8)*(2x+1)/(2x + 5) = e^(-8)`

Hence, evaluating the given limit, using special limit, yields `lim_(x->oo) ((2x - 3)/(2x + 5))^(2x + 1) = 1/(e^8).`

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