sin2x1x趋近1

俞养满3787lim x^2 sin1/x,也是求极限,x趋于0 -
焦肢玉13145517790 ______[答案] 1/x趋于无穷 sin1/x属于[-1,1] 即有界 无穷小乘有界是无穷小 所以极限=0

俞养满3787[sin(x^2)sin(1/x)]/[ln(1+2x)]的极限(X趋向于0) -
焦肢玉13145517790 ______[答案] 此式是乘除一体的,所以可以用这个方法:sin(x^2)在X趋向于0时为x^2,sin(1/x)在X趋向于0时为sin(1/x)也是小于等于1大于等于-1;ln(1+2x)在X趋向于0时为2x;所以此式的极限为0

俞养满3787高数求极限.x趋向于零,lim(sin^2x - x^2cos^2x)/x^2sin^2x ,高数求极限.x趋向于零,lim(sin^2x - x^2cos^2x)/x^2sin^2x , -
焦肢玉13145517790 ______[答案] sinx=x-1/6x^3+o(x^3) 则sin^2x=x^2-1/3x^4+o(x^4) cosx=1-x^2/2+o(x^4) cos^2x=1-x^2+o(x^4) 那么分子就是sin^2x-x^2cos^2x=x^2-1/3x^4+o(x^4)-[x^2-x^4+o(x^4)]=2/3x^4+o(x^4) 所以结果是2/3 或者sin^2x-x^2cos^2x=(sinx+xcosx)(sinx-xcosx) 那么原式=lim...

俞养满3787高等数学.sin(x+x1) - sinx=2sin(x1/2)cos(x+x1/2),这是怎么转换的? -
焦肢玉13145517790 ______ 根据和差化积公式: sinA - sinB = 2cos((A+B)/2)sin((A-B)/2) A = x + x1 B = x (A + B)/2 = (2x + x1)/2 = x + x1/2 (A - B)/2 = (x + x1 - x)/2 = x1/2 因此,sin(x + x1) - sinx = 2sin(x1/2)cos(x + x1/2)

俞养满3787当x趋向于0,求lim{ln(1+2sin^2x)/1 - cosx} -
焦肢玉13145517790 ______[答案] lim{ln(1+2sin^2x)/1-cosx} =lim (2sin^2x)/(x^2/2) =lim 2x^2/(x^2/2) =4

俞养满3787limx→∞x*sin(2x/(x^2+1))的极限 -
焦肢玉13145517790 ______[答案] x*sin(2/(x+1))

俞养满3787解下列方程:2sin^2x=1,0不是2x...是2sin平方x=1,0 -
焦肢玉13145517790 ______[答案] 2sin²x=1 sin²x=1/2 sinx=±√2/2 所以 x=π/4或x=3π/4或x=5π/4或x=7π/4

俞养满3787x趋于无穷,lim(2x^2+1/x+2)sin2/x的值是多少 -
焦肢玉13145517790 ______[答案] 显然x趋于无穷的时候, 2/x趋于0, 而在A趋于0的时候,sinA等价于A, 即在这里sin2/x等价于2/x 所以 lim(x->∞)(2x^2+1/x+2)*sin2/x =lim(x->∞)(2x^2+1/x+2)* 2/x =lim(x->∞) 4x+2/x^2 +4/x 而x趋于∞时,4x趋于∞,而2/x^2 +4/x趋于0 所以 此极限为∞

俞养满3787lim tanx - sinx/sin^2x x趋近于0,x的极限 -
焦肢玉13145517790 ______[答案] (tanx-sinx)/sin²x =(sinx/cosx-sinx)/sin²x =(1/cosx-1)/sinx =(1-cosx)/sinacosx x趋于0 所以1-cosx~x²/2 sinx~x 所以原式=lim(x²/2)/xcosx =limx/(2cosx) =0

俞养满3787lim 1 - cos4x/2sin^2x+xtan^2x x趋近于0 求极限 -
焦肢玉13145517790 ______[答案] 因为1-cos4x/2sin^2x+xtan^2x =1-(1-2sin^2x)/2sin^2x+sin^2x/cos^2x =2sin^2x/2sin^2x+sin^2x/cos^2x =2/1+1/cos^2x 所以原式=2/2=1