Operiamo la sostituzione y= 2x:
lim
x
→
0
(
1
+
2
x
)
5
−
1
x
=
lim
y
→
0
(
1
+
y
)
5
−
1
y
/
2
=
2
⋅
lim
y
→
0
(
1
+
y
)
5
−
1
y
lim from{x toward 0 } {{ (1+2x)^5 -1 } over { x } } = lim from{y toward 0 } {{ (1+y)^5 -1 } over { y/2 } }= 2 cdot lim from{x toward 0 } {{ (1+y)^5 -1 } over { y } }
Ricordando il limite notevole:
lim
x
→
0
(
1
+
x
)
k
−
1
x
=
k
lim from{x toward 0 } {{ (1+x)^k -1 } over { x } }=k
Si ha:
lim
x
→
0
(
1
+
2
x
)
5
−
1
x
=
2
⋅
lim
y
→
0
(
1
+
y
)
5
−
1
y
=
2
⋅
5
=
10
lim from{x toward 0 } {{ (1+2x)^5 -1 } over { x } } = 2 cdot lim from{x toward 0 } {{ (1+y)^5 -1 } over { y } }= 2 cdot 5 = 10