أولا عندما يؤول x إلى 0 و x > 0

0 < x < 1
0 < x < x^(1/3) < 1
0 < sin(x) < sin(x^(1/3))
cos(x) < 1
0 < 1 - cos(x)
sin(x)/(1-cos(x)) < sin(x^(1/3))/(1-cos(x))
lim sin(x)/(1-cos(x)) < lim sin(x^(1/3))/(1-cos(x))

نحدد نهاية ((sin(x) / (1-cos(x عندما يؤول x إلى 0 موجب

lim sin(x)/(1-cos(x))
= lim sin(x)(1+cos(x))/(1-cos²(x))
= lim sin(x)(1+cos(x))/sin²(x)
= lim (1+cos(x))/sin(x)

= lim cos(x)(1/cos(x)+1)/sin(x)
= lim (1/cos(x)+1)/tan(x)
= lim (1/1+1)/0+ = 2/0+ = +inf

ومن التأطير الأول

lim sin(x)/(1-cos(x)) < lim sin(x^(1/3))/(1-cos(x))
+inf < lim sin(x^(1/3))/(1-cos(x))
lim sin(x^(1/3))/(1-cos(x)) = +inf

ثانيا عندما يؤول x إلى 0 و x < 0

نستنتجه من النتيجة السابقة

x < 0
lim sin((-x)^(1/3))/(1-cos(-x)) = +inf
lim sin(-x^(1/3))/(1-cos(x)) = +inf
lim -sin(x^(1/3))/(1-cos(x)) = +inf
lim sin(x^(1/3))/(1-cos(x)) = -inf