Stimulated by Jakobsen and Pellegrini's Love the Sin: Sexual Regulation and the Limits of Religious Tolerance, this essay is an effort to extend the value of their contribution by including a psychoanalytic perspective within its purview. My intent is to bring psychoanalytic readers into this ...
Homosexuals, Heretics, and the Practice of Freedom Commentary on Love the Sin: Sexual Regulation and the Limits of Religious Tolerance by Janet R. Jakobsen and Ann PelligriniHere I construe Janet R. Jakobsen and Ann Pellegrini's proposal to protect freedom of sexual expression among consenting ...
**1. Simplify the expression:** We can simplify the expression using trigonometric identities: * tan x = sin x / cos x * sec x = 1 / cos x Substituting these identities, we get: ``` lim (x² sin x (sin x / cos x)) x→0 (4 cos x + 4 (1 / cos x) - 8) `...
所以,lim_(x → 0) (tan x - sin x)/(x sin x (1 - cos x)) = 0。 该极限问题可以通过洛必达法则求解。由于分子分母在x趋近于0时都趋近于0,满足洛必达法则的条件。对分子分母分别求导,得到新的极限表达式。最终计算得到极限值为0。反馈 收藏 ...
12. 计算\lim\limits_{x{\to}0}\frac{x-{\tan}x}{{\sin}x(1-\sqrt{{\cos}x}} 相关知识点: 试题来源: 解析limlimits_(x→0)(x-(tan)x)/((sin)x(1-√((cos)x)) = -4/3 该极限可以使用洛必达法则和泰勒展开式求解。由于分子和分母在 x → 0 时都趋于零,我们可以使用...
首先,将表达式化简为 limlimits_(x→0)(tan x - x)/(x-sin x)。由于当 x 趋近于 0 时,分子和分母都趋近于 0,因此可以应用洛必达法则,对分子和分母分别求导,得到 limlimits_(x→0)(sec^2 x - 1)/(1-cos x)。再次应用洛必达法则,得到 limlimits_(x→0)(2sec^2 x tan x)/(...
百度试题 结果1 题目lim limits _(x→ 0) (tan x-sin x)(x^3) 相关知识点: 试题来源: 解析 12反馈 收藏
lim_{(x, y) to (3, 2)} (x^2 y - y^3 + cos (pi x)) Evaluate the limit \lim_{x \to 0} \frac{\sin (x)}{x + \tan (x)} Evaluate the limit. lim_x to 0 (2 x^2 - 3 x + 4 / x + 5 x - 4 / x) Evaluate the limit. \lim_(x,y)} \rightarrow {...
题目4: 利用洛必达法则,对分子分母分别求导,并利用重要极限 lim_(x → 0) (sin x)/x = 1,最终得到结果为 1/2。 题目5: 利用重要极限 lim_(x → 0) (1 1/x)^x = e,将原式转化为指数形式,并利用洛必达法则求解,最终得到结果为 √[3](abc)。反馈...
(7)\lim\limits_{x{\to}0}\frac{x{\tan}x{\sin}\frac{1}{x}}{{\tan}5x}; 相关知识点: 试题来源: 解析 极限不存在。 该极限可以通过洛必达法则求解。由于分子和分母在x趋近于0时都趋近于0,因此可以对分子和分母分别求导。经过两次洛必达法则后,发现极限表达式中包含sin1/x,该项在x趋近...