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ayase252

ayase252

V2EX 第 241507 号会员,加入于 2017-07-20 01:10:06 +08:00
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ayase252 最近回复了
你买不推荐,我买我不买
11 天前
回复了 gtese 创建的主题 问与答 鬼压床,你们怎么解?
享受这种想动动不了的感觉
17 天前
回复了 ShikiSuen 创建的主题 Apple 新 iPad pro 记得买 1TB 以上的
如果他能够满足哪怕只有 80% 的开发生产力,我闭眼买。
但是呵呵
18 天前
回复了 dilidilid 创建的主题 Apple 现在 iPad Pro 的价格有点太夸张了吧
笑稀了,11500 买 iPad ,库克你也做的出来
30 天前
回复了 Persimmon08 创建的主题 OpenAI 这个问题,大家的 GPT 都能答对吗?
gemini pro

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.
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