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Textbook solution for single variable calculus: Early transcendentals 8th edition james stewart chapter 2. 6 problem 35e. We're not allowed to take the logs or exp(. ) because we haven't learned that yet. Evaluate the limits at infinity.
Lim x → ± ∞ x 2 1 − x 2 = lim x → ± ∞ 1 1 x 2 − 1 = −1. Therefore, f has a horizontal asymptote of y = −1 as x. To answer this question, you need to know that lim x→+ ∞ ex = + ∞ and lim x→+∞ arctanx = π 2 from the stuy of ex (see exponential functions ) and of arctanx (see inverse cosine and inverse tangent ). So, as x → ∞, ex → ∞ so that, letting t = ex we have. Lim x→∞ arctan(ex) = lim t→ ∞ arctan(t) = π 2. Textbook solution for single variable calculus: Concepts and contexts,… 4th edition james stewart chapter 3. 6 problem 39e. 本站是关于蜂蜜的介绍网站， 资源文档来自于网友推荐，仅作参考作用。 若有相关事宜，请联系电邮 © 蜂蜜营养网蜂蜜营养网 What is the limit of e to infinity?
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