IGN发布了《极限竞速》新作的Demo试玩视频，捕捉自Xbox Series X，4K、60FPS。开发商Turn 10指出Demo仅是发售前版本，并不代表最终游戏的品质。

“Turn 10团队将带我们体验Xbox Series X《极限竞速》的游戏玩法，并解释新系统将如何在比赛中解锁新的汽车组件。这款4K《极限竞速》实机是在Xbox Series X上录制，并展示了我们对这款模拟赛车游戏在今年10月发售时的期待。”

视频：

《极限竞速》新作将在首发时提供超过500辆真实世界的汽车，游戏将包含超过800个独特的升级和自定义修改。不仅如此，游戏开始将有20种环境，具有多种赛道布局供玩家掌控。

Turn10工作室成功实现了10倍更真实的细节呈现，包括高度详细的植被和过度生长。此外，《极限竞速》新作将包括数以万计的完全动画化的3D观众。

《极限竞速》新作将于10月10日发售。

视频画面：

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甄采惠885求下列极限li m((1 - x^2/2)^2/3 - 1)/(xln(1+x)) X趋向于0 -
鞠路米17787745702 ______[答案] (1-x^2/2)^(2/3)-1 2/3·(-x^2/2) = -x^2/3 xln(1+x)~x·x=x^2 =lim(-1/3x^2/x^2) = -1/3

甄采惠885为什么 lim(x趋向于0)(In(1+x)/x)=lim(x趋向于0)In(1+x)^(1/x) ? -
鞠路米17787745702 ______ lim(x趋向于0)(In(1+x)/x)=lim(x趋向于0)In(1+x)^(1/x) 这个是从公式mlnn=lnn^m来的啊 mlnn=lnn^m可以通过lna+lnb=lnab推出

甄采惠885利用取对数的方法 求(1+1/x^2)^x的极限 -
鞠路米17787745702 ______ lnt=ln(1+x^2)/x ~x 当x趋近于0,lnt~0,所以t~1 极限1

甄采惠885当x趋近于0时,lim(3sinx+x^2 cos1/x)'/x的极限为什么不存在?' -
鞠路米17787745702 ______ 当x趋近于0时, lim(3sinx+x^2 cos1/x)'/x =lim(3cosx+2xcos1/x+sin1/x)/x 当x趋近于0时,sin1/x极限不存在,3cosx+2xcos1/x极限存在为3 所以3cosx+2xcos1/x+sin1/x不存在,因为当x趋近于0时,x的极限存在为0 所以lim(3cosx+2xcos1/x+sin1/x)/x的极限不存在 可以用反证法证明!!

甄采惠885lim(1/ln(x+1) - 1/x) x趋于零 用洛必达法则 -
鞠路米17787745702 ______ 解:原式=[x-ln(x+1)]/xln(x+1) =[x-ln(x+1)]/x^2 【 ln(x+1)和X是等价无穷小 在x趋于0时】 =[1-1/(x+1)]/2x 【0/0型洛必达法则】 =x/2x(x+1) =1/2

甄采惠885利用等价无穷小替换 lim 的x趋向于0 ln(1+x)/x是多少?(要过程)谢谢~~ -
鞠路米17787745702 ______ x趋向于0时 ln(1+x)与x就等价 所以:原式=lim x/x =lim 1 =1

甄采惠885当x趋向于0,(e^x+sinx - --2x - 1)/[xln(1+x)]的极限-- -
鞠路米17787745702 ______ 看不清题目,但最通用的方法是用麦克劳林展开:当x→0时,e^x → 1 + x + x² + o(x²) 至于需要展开到几阶,要根据具体题目,估计这道题需要展开为3项 sinx→ x - x³/6 + o(x³) ln(1 + x) → x - x²/2 + o(x²) lim (e^x + sinx - 2x - 1) / [xln(1+x)] =...

甄采惠885求Xln1/X极限、x趋于0.为什么将后面换成1/X一1求与洛必达法则求结果不一样啊?这里ln1/x不能写成1/x一1吗极限X(1/x一1)=0错的为什么Ln1/x与1/x—1不等... -
鞠路米17787745702 ______[答案] 不等价的 等价的是 ln(1+x)=x x趋于0的时候 你俩边都看做e的幂 再积分 就显然不等价了吧~

甄采惠885n趋于无穷大,(1+x)/(1+x^2n)的极限是多少?过程 -
鞠路米17787745702 ______ 1,当/X/>1时,x^2n趋向正无穷,(1+x^2n)趋向正无穷,,(1+x)/(1+x^2n)趋向于0,极限是02,当/X/<1时,x^2n趋向0,(1+x^2n)趋向于1,,(1...

甄采惠885求极限X趋近于0+[Xln(1+3X)]/[(1 - cos2x)^2]和X趋近于0 [根号(1+x) - 三次根号(1+2x^2)]/ln(1+3x) -
鞠路米17787745702 ______[答案] (1)只要注意到ln(1+x)~x(x→0),sinx~x(x→0),以及cos倍角公式:1-cos2x=2(sinx)^2 容易知道极限趋向于+∞ (2)只要知道(1+x)^a~ax(x→0)就容易知道极限=lim(x→0)(1/2*x-2/3*x^2)/3x=1/6