提问哪个大模型？一周就用了 50w……

想着使用 java 开发个东西，这才是提高的好方法，不然光背八股文和看培训课程，那始终是纸上谈兵

快递外卖，外卖保安，你上白班，我上夜班

@iamtuzi3333 还有一种就是工作用不到，用不到自然不想学了

说明压力不够，就好像有了老婆孩子，再不想工作也待去挣钱养家

看完了老哥写的，职业生涯提不了有建设性的，祝老哥前途一路顺风，

现在的环境不都是等着毕业吗？难道还真能干到 65 岁？

@kangfenmao 老哥，有没有网页端啊？

看微信，看 B 站，看漫画，听歌，足够了

@austinxuyan 日区吧，最近日元贬值

刚需需要房子住，别管涨跌买，不是刚需就算了

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

索尼、摩托罗拉、Zenfone 、谷歌，一加已经是 colorOS 了，

有消息了吗？

@tcpdump +1 ，我的也是，你的是美国手机号吗？

靠近香港去香港，离香港太远去某宝

外卖、快递、导航

恭喜