求极限,大学微积分

lim (x+e∧x)∧1/x
x→0
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曼斯拉乌

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lim_(x->0) (e^x + x)^(1/x) = lim_(x->0) e^(log((e^x + x)^(1/x))):
lim_(x->0) e^(log((e^x + x)^(1/x)))
e^(log((e^x + x)^(1/x))) = exp((log(e^x + x))/x):
lim_(x->0) exp((log(e^x + x))/x)
lim_(x->0) e^((log(e^x + x))/x) = e^(lim_(x->0) (log(e^x + x))/x):
e^(lim_(x->0) (log(e^x + x))/x)
Write log(e^x + x) as log(e^x) + log(1 + e^(-x) x):
e^(lim_(x->0) log(e^x) + log(1 + e^(-x) x)/x)
(log(e^x) + log(1 + e^(-x) x))/x = (log(1 + e^(-x) x) + x)/x:
e^(lim_(x->0) (log(1 + e^(-x) x) + x)/x)
Applying l'Hôpital's rule, we get that
lim_(x->0) (x + log(1 + e^(-x) x))/x | = | lim_(x->0) ( d/( dx)(x + log(1 + e^(-x) x)))/(( dx)/( dx))
| = | lim_(x->0) (1 + (e^(-x) - e^(-x) x)/(1 + e^(-x) x))/1
| = | lim_(x->0) (1 - (x - 1)/(e^x + x))
e^lim_(x->0) (1 - (x - 1)/(e^x + x))
lim_(x->0) (1 - (x - 1)/(e^x + x)) = 1 - (0 - 1)/(e^0 + 0) = 2:
Answer: e^2

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