笑稀了，11500 买 iPad ，库克你也做的出来

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Here's how to find the limits of x*ln(1+1/x) as x approaches 0 and positive infinity:

Case 1: x approaches 0

We cannot directly substitute x = 0 into the expression, as it results in an indeterminate form. Let's use L'Hôpital's rule:

Rewrite the expression:
x*ln(1+1/x) = ln(1+1/x) / (1/x)
Now the form is suitable for L'Hôpital's rule (both numerator and denominator approach 0 as x approaches 0).

Apply L'Hôpital's Rule:
lim (x->0) [ln(1+1/x) / (1/x)] = lim (x->0) [d/dx(ln(1+1/x)) / d/dx(1/x)]
= lim (x->0) [-1/(x+x^2) / (-1/x^2)]
= lim (x->0) [x / (1+x)] = 0

Therefore, the limit of x*ln(1+1/x) as x approaches 0 is 0.

Case 2: x approaches positive infinity

Rewrite the expression:
x*ln(1+1/x) = ln(1 + 1/x) / (1/x)

Again, this is an indeterminate form (both numerator and denominator approach 0 as x approaches infinity).

Apply L'Hôpital's Rule:
lim (x-> ∞) [ln(1 + 1/x) / (1/x)] = lim (x-> ∞) [d/dx(ln(1 + 1/x)) / d/dx(1/x)]
= lim (x-> ∞) [-1/(x+x^2) / (-1/x^2)]
= lim (x-> ∞) [x / (x+1)]

= 1 (dividing both numerator and denominator by x)

Therefore, the limit of x*ln(1+1/x) as x approaches positive infinity is 1.

大城市租，小城市交通不便 798

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正经科研不看中文 paper 的

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之前回都是全价，今年来回都打折了