`lim_(x->oo) x^(1/x)` 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...

`lim_(x->oo) x^(1/x)` 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 61 - 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^o` . You need to use the logarithm technique, such that:

`f(x) = (x)^(1/x)`

You need to take logarithms both sides, such that:

`ln f(x) = ln (x)^(1/x))`

Using the property of logarithms yields:

`ln f(x) = (1/x)*ln x`

`ln f(x) = (ln x)/x`

You need to evaluate the limit:

`lim_(x->oo) ln f(x) = lim_(x->oo) (ln x)/x = oo/oo`

You need to use L'Hospital theorem:

`lim_(x->oo) (ln x)/x = lim_(x->oo) ((ln x)')/(x') `

`lim_(x->oo) ((ln x)')/(x') = lim_(x->oo) (1/x)/1`

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

Hence, `lim_(x->oo) ln f(x) = 0` , such that `lim_(x->oo) f(x) = e^0 = 1`

Hence, evaluating the given limit, using l'Hospital rule and logarithm technique yields `lim_(x->oo) (x)^(1/x) = e^0 = 1` .

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scisser | (Level 3) Honors

Posted on

Plug in `oo ` everywhere you x

`lim_(x-gtoo)x^(1/oo)`

1 divided by a really big number is just 0

Therefore,

`lim_(x-gtoo)x^0`

`=1`

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