Question

三角函数极限

x趋于0+ 1-cos(根号x)/sin(2x)
x趋于无穷 分子 a*根号下x^4+5 分母 bx^2+cx*sin(x^2+1)
要求不要用洛必达法则做

Answer 1

x趋于0+ 1-cos(根号x)/sin(2x)
=x趋于0+ [1-cos(根号x)]/sin(2x)
=x趋于0+ [sin(根号x)*1/2*x^(-1/2)]/2cos(2x)
=x趋于0+ [1/2 *sin(根号x)/根号x]/2cos(2x)
=x趋于0+[sin(根号x)]/[4(根号x)*cos(2x)]
=x趋于0+[sin(根号x)]/[4(根号x)]
=x趋于0+[cos(根号)*1/2*x^(-1/2)]/[4*1/2*x^(-1/2)]
=x趋于0+[cos(根号x)]/4
=cos0/4
=1/4
x趋于无穷 分子 a*根号下x^4+5 分母 bx^2+cx*sin(x^2+1)

=x趋于无穷 ( a*根号下x^4+5 )/[ bx^2+cx*sin(x^2+1)]
=x趋于无穷[a*1/2*(x^4+5)^(-1/2) *4x^3]/[2bx+c*sin(x^2+1)
+cx*cos(x^2+1) *2x]
=x趋于无穷[2ax^3/(x^4+5)^(1/2)]/[2bx+c*sin(x^2+1)+2cx^2cos(x^2+1)]
=x趋于无穷[2a/(1+5/x^4)^(1/2)]/[2b+c/x *sin(x^2+1)+2cxcos(x^2+1)]
=x趋于无穷[(2a/x)/(1+5/x^4)^(1/2)]/[2b/x+c/x^2 *sin(x^2+1)+2ccos(x^2+1)
=(x趋于无穷(0/1)/(0+0+2ccos(x^2+1)
=0