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h(x) = x -1 + \frac{1+ ln {}^(2) (x) }{x} \displaystyle \lim_(x\to0) h(x)= \: ? \\ \displaystyle \lim_(x\to \infty ) h(x)= \: ? Apply L'Hôpital's Rule if possible
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Jun 22, 2023
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Apply L'Hôpital's Rule if possible
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Answer:
Explanation:
= +∞ + 0
= +∞
= -1 + +∞
= +∞
Anthoney
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Jun 27, 2023
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