(1)lim(x趋向0)[(1+x)^0.5+(1-x)^0.5-2]/x^2 (2)lim(n趋向无穷)sin[[(n^2+1)^0.5]π](1)lim(x趋向0)[(1+x)^0.5+(1-x)^0.5-2]/x^2 (2)lim(n趋向无穷)sin[[(n^2+1)^0.5]π]

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(1)lim(x趋向0)[(1+x)^0.5+(1-x)^0.5-2]/x^2 (2)lim(n趋向无穷)sin[[(n^2+1)^0.5]π](1)lim(x趋向0)[(1+x)^0.5+(1-x)^0.5-2]/x^2 (2)lim(n趋向无穷)sin[[(n^2+1)^0.5]π]

(1)lim(x趋向0)[(1+x)^0.5+(1-x)^0.5-2]/x^2 (2)lim(n趋向无穷)sin[[(n^2+1)^0.5]π](1)lim(x趋向0)[(1+x)^0.5+(1-x)^0.5-2]/x^2 (2)lim(n趋向无穷)sin[[(n^2+1)^0.5]π]
(1)lim(x趋向0)[(1+x)^0.5+(1-x)^0.5-2]/x^2 (2)lim(n趋向无穷)sin[[(n^2+1)^0.5]π]
(1)lim(x趋向0)[(1+x)^0.5+(1-x)^0.5-2]/x^2
(2)lim(n趋向无穷)sin[[(n^2+1)^0.5]π]

(1)lim(x趋向0)[(1+x)^0.5+(1-x)^0.5-2]/x^2 (2)lim(n趋向无穷)sin[[(n^2+1)^0.5]π](1)lim(x趋向0)[(1+x)^0.5+(1-x)^0.5-2]/x^2 (2)lim(n趋向无穷)sin[[(n^2+1)^0.5]π]
我只知道问题一 将x=0代入 为0/0型 用罗比达法则 lim(x趋向0){1/2[(1+x)^-0.5-(1-x)^-0.5]/2x}
= lim(x趋向0){[-1/4(1+x)^-1.5+1/4(1-x)^-1.5]/2}
将x=0代入得 =0
问题二 n趋向无穷==>[(n^2+1)^0.5]π=nπ
即lim(n趋向无穷)sin[[(n^2+1)^0.5]π]=lim(n趋向无穷)sin nπ 因为函数y=sinx为周期函数 所以 sin nπ n—>无穷 极限不存在

1.x趋向0 lim (cosx-1)/x 2.x趋向0 lim sinx/x lim(1-cosx)x趋向0, lim(x趋向0) 1-cos2x/xsinx x趋向0 lim [ ln (1-x) / (e ^ x-1 ) ] 计算极限 lim(1-x)^1/x x趋向0 lim(1+e^1/x)^x,x趋向于+0 lim x趋向0 (1+cos^2(1/x))^x lim(1-2x)^1/x x趋向于0 lim(x^2+1)/(2x^+1) x趋向0 求极限 lim x趋向0(x+ex)1/x lim(ln(1+x)/x) x趋向0 lim(x趋向0)ln(1+sin x)/x^2 求lim x趋向于0(e^x-x-1) x趋向0 lim(cot x- 1/x)=? x趋向于0 lim loga(1+x)/x= lim(x趋向于0)(x*cotx-1)/(x^2) x趋向0+时求lim(x^x-1)*lnx lim(1+3x)^3/x x趋向于0