Esprimiamo la radice come una potenza razionale:
lim
x
→
0
x
+
1
5
−
1
5
x
=
1
5
⋅
lim
x
→
0
(
x
+
1
)
1
5
−
1
x
lim from{x toward 0 } {{ nroot{5}{x+1} -1 } over { 5x } } = 1 over 5 cdot lim from{x toward 0 } {{ (x+1)^{1 over 5} -1 } over {x } }
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
x
+
1
5
−
1
5
x
=
1
5
⋅
lim
x
→
0
(
x
+
1
)
1
5
−
1
x
=
1
5
⋅
1
5
=
1
25
lim from{x toward 0 } {{ nroot{5}{x+1} -1 } over { 5x } } = 1 over 5 cdot lim from{x toward 0 } {{ (x+1)^{1 over 5} -1 } over {x } }= {1 over 5} cdot {1 over 5}= 1 over 25