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| V 站有办理过 Ultra Mobile PayGo 手机卡的老哥吗？ 新手求助 • Wkj1998 • 48 天前 • 最后回复来自 zhhmax | 13 |
| 除了转运还有其他办法能寄海淘的商品吗？ 问与答 • Wkj1998 • 2022-09-20 15:52:18 PM • 最后回复来自 dangyuluo | 2 |
| 老哥们有没有啥方法能开这个 Google one Google • Wkj1998 • 2022-09-11 07:45:00 AM • 最后回复来自 fatanstic4 | 4 |
| 收个优酷年卡/半年卡 二手交易 • Wkj1998 • 2020-07-13 18:11:16 PM • 最后回复来自 coobin | 8 |
Wkj1998 最近回复了
22 小时 22 分钟前 回复了 Persimmon08 创建的主题 › OpenAI › 这个问题，大家的 GPT 都能答对吗？ |
Sure, I'd be happy to help you with that advanced math problem!
To find the limit of y = x ln(1 + 1/x) as x tends to 0 and positive infinity, we need to apply the appropriate limit laws and algebraic manipulations.
Limit as x tends to 0:
Let's first simplify the expression by applying the rule: lim(x→0) x ln(1 + 1/x) = lim(x→0) ln(1 + 1/x) / (1/x)
Now, we can use the limit laws and algebraic manipulations:
lim(x→0) ln(1 + 1/x) / (1/x) = lim(x→0) [ln(1 + 1/x)] / [(1/x)]
= lim(x→0) [ln(x + 1) - ln(x)] / [(x + 1 - x)/x]
= lim(x→0) [ln(x + 1) - ln(x)] / [1/x]
= lim(x→0) [(d/dx) ln(x)] (Using L'Hospital's rule)
= lim(x→0) 1/x
= ∞
Limit as x tends to positive infinity:
lim(x→∞) x ln(1 + 1/x) = lim(x→∞) ln(1 + 1/x) / (1/x)
= lim(x→∞) [ln(x + 1) - ln(x)] / [(x + 1 - x)/x]
= lim(x→∞) [ln(x + 1) - ln(x)] / [1/x]
= lim(x→∞) [(d/dx) ln(x)] (Using L'Hospital's rule)
= lim(x→∞) 1/x
= 0
Therefore, the limit of y = x ln(1 + 1/x) as x tends to 0 is positive infinity (∞), and the limit as x tends to positive infinity is 0.
Claude Sonnet
To find the limit of y = x ln(1 + 1/x) as x tends to 0 and positive infinity, we need to apply the appropriate limit laws and algebraic manipulations.
Limit as x tends to 0:
Let's first simplify the expression by applying the rule: lim(x→0) x ln(1 + 1/x) = lim(x→0) ln(1 + 1/x) / (1/x)
Now, we can use the limit laws and algebraic manipulations:
lim(x→0) ln(1 + 1/x) / (1/x) = lim(x→0) [ln(1 + 1/x)] / [(1/x)]
= lim(x→0) [ln(x + 1) - ln(x)] / [(x + 1 - x)/x]
= lim(x→0) [ln(x + 1) - ln(x)] / [1/x]
= lim(x→0) [(d/dx) ln(x)] (Using L'Hospital's rule)
= lim(x→0) 1/x
= ∞
Limit as x tends to positive infinity:
lim(x→∞) x ln(1 + 1/x) = lim(x→∞) ln(1 + 1/x) / (1/x)
= lim(x→∞) [ln(x + 1) - ln(x)] / [(x + 1 - x)/x]
= lim(x→∞) [ln(x + 1) - ln(x)] / [1/x]
= lim(x→∞) [(d/dx) ln(x)] (Using L'Hospital's rule)
= lim(x→∞) 1/x
= 0
Therefore, the limit of y = x ln(1 + 1/x) as x tends to 0 is positive infinity (∞), and the limit as x tends to positive infinity is 0.
Claude Sonnet
31 天前 回复了 louettagfh 创建的主题 › Android › 2024 年了，类原生 Android 手机还有什么推荐? |
索尼、摩托罗拉、Zenfone 、谷歌，一加已经是 colorOS 了，
51 天前 回复了 uedkkk 创建的主题 › 职场话题 › 技术面却聊人生，是挂了吗？ |
有消息了吗？
58 天前 回复了 puputeju 创建的主题 › OpenAI › 成功注册 Claude3 且不被封号 |
@tcpdump +1 ，我的也是，你的是美国手机号吗？
63 天前 回复了 balabalaguguji 创建的主题 › 程序员 › S24 港版你们去哪购买？ |
靠近香港去香港，离香港太远去某宝
77 天前 回复了 996bujiaban 创建的主题 › 算法 › 人类目前算法领域最高的结晶,是什么? |
外卖、快递、导航
80 天前 回复了 ggggz 创建的主题 › 程序员 › 做的单词翻译 API 收到第一个付费订阅了！ 20 美刀！ |
恭喜
83 天前 回复了 justincnn 创建的主题 › Google › gemini 的 app 无法使用，大家有碰到么？ |
也有可能检测到不是美国 IP
83 天前 回复了 justincnn 创建的主题 › Google › gemini 的 app 无法使用，大家有碰到么？ |
我也是这样的，会不会必须要谷歌手机才可以？
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