`lim_(x->0) (1 - 2x)^(1/x)` Find the limit. Use l’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it.

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Chapter 4, 4.4 - Problem 57 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to evaluate the given limit, hence, you need to replace 0 for x, such that:

`lim_(x->0)(1 - 2x)^(1/x) = (1 - 0)^(1/0) = 1^oo`

You may use special limit `lim_(x->0) (1 + x)^(1/x) = e` , instead of using l'Hospital's rule, such that:

`lim_(x->0)((1 + (-2x))^(1/(-2x)))^((-2x)/x) = e^lim_(x->0)((-2x)/x)`

Evaluate separately `lim_(x->0)((-2x)/x) = lim_(x->0) (-2) = -2.`

`lim_(x->0)((1 + (-2x))^(1/(-2x)))^((-2x)/x) = e^(-2)`

Hence, evaluating the limit, yields `lim_(x->0)(1 - 2x)^(1/x) = 1/(e^2).`

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