Dividiamo numeratore e denominatore con 2x:
lim
x
→
0
sin
2
x
x
cos
x
=
lim
x
→
0
sin
2
x
2
x
x
cos
x
2
x
=
2
⋅
lim
x
→
0
sin
2
x
2
x
lim
x
→
0
cos
x
lim from{x toward 0 } {{ sin 2x } over { x cos x } }=lim from{x toward 0 } {{ {sin 2x} over {2x} } over { {x cos x} over {2x} } }= 2 cdot {{lim from{x toward 0 }{ {sin 2x} over {2x}} } over { lim from{x toward 0 } { cos x} } }
Ricordiamo ora il limite notevole:
lim
y
→
0
sin
y
y
=
1
lim from{y toward 0 } { {sin y} over y}=1
e calcoliamo il limite:
lim
x
→
0
sin
2
x
x
cos
x
=
2
⋅
lim
x
→
0
sin
2
x
2
x
lim
x
→
0
cos
x
=
2
⋅
1
1
=
2
lim from{x toward 0 } {{ sin 2x } over { x cos x } }=2 cdot {{lim from{x toward 0 }{ {sin 2x} over {2x}} } over { lim from{x toward 0 } { cos x} } }= 2 cdot {1 over 1 } = 2