Here is how we can find the limit of y = x ln(1 + 1/x) as x approaches 0 and positive infinity:
Limit as x → 0⁺
As x approaches 0 from the right side, 1/x approaches positive infinity. So we can rewrite the limit:
lim x→0⁺ x ln(1 + 1/x) = lim x→0⁺ x ln(1/x)
Now, using the properties of logarithms, ln(1/x) = -ln(x). So:
= lim x→0⁺ -x ln(x)
This is an indeterminate form of type 0 ⋅ ∞. We can use L'Hôpital's rule to evaluate it:
= lim x→0⁺ -ln(x) - 1 = -∞
Therefore, lim x→0⁺ x ln(1 + 1/x) = -∞
Limit as x → +∞
As x approaches positive infinity, 1/x approaches 0. Using the fact that ln(1+u) ≈ u for small u:
lim x→+∞ x ln(1 + 1/x) ≈ lim x→+∞ x ⋅ 1/x = lim x→+∞ 1 = 1
Therefore, lim x→+∞ x ln(1 + 1/x) = 1
In summary:
As x → 0⁺, y = x ln(1 + 1/x) → -∞
As x → +∞, y = x ln(1 + 1/x) → 1
claude3 opus 很准。
claude-3-opus-20240229
当 x 趋近于 0 时，y 的极限为 0 ；当 x 趋近于正无穷时，y 的极限为 1 。
claude-3-haiku-20240307
当 x 趋近于 0 和正无穷时,函数 y = x ln(1+1/x) 的极限都是 1
claude-3-sonnet-20240229
当 x 趋近于 0 和正无穷时,极限 y = x ln(1+1/x) 的值均为 1 。
貌似只有 opus 是对的
