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6 天前 回复了 eedwinhei 创建的主题 › 问与答 › 怎么看待最近国内房地产政策频发？买不买？ |
你买不推荐，我买我不买
9 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
15 天前 回复了 gsy20050126 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
但是呵呵
16 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
28 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.
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.
35 天前 回复了 ashin 创建的主题 › 分享创造 › 写了一个简化版退休金在线估算器，算算你退休时可以领多少养老金 |
2050 年想 60 退休，想多了
62 天前 回复了 drymonfidelia 创建的主题 › 问与答 › IPv4 总量也是固定、不可能增发的，为什么没有涨到 BTC 这种价格？ |
你并不拥有 IP 地址，运营商说收回就收回哪来的价值
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