# `lim_(x->oo) (e^x + 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,...

`lim_(x->oo) (e^x + 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.

### Textbook Question

Chapter 4, 4.4 - Problem 62 - 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 following technique, such that:

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

You need to take logarithms both sides, such that:

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

Using the property of logarithms yields:

`ln f(x) = (1/x)*ln ((e^x + x))`

`ln f(x) = (ln ((e^x + x)))/x`

You need to evaluate the limit:

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

You need to use L'Hospital theorem:

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

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

`lim_(x->oo) ((e^x + x)')/(e^x + x) = lim_(x->oo) (e^x + 1)/(e^x + x)`

You need to factor out `e^x` such that:

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

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

Since `lim_(x->oo)1/(e^x) = 0` and `lim_(x->oo) x/(e^x) = ` 0 yields:

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

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

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