Si risolve la forma indeterminata di questo limite dividendo numeratore e denominatore con la variabile x:
lim
x
→
+
∞
1
+
x
2
4
x
−
3
=
lim
x
→
+
∞
1
+
x
2
x
4
x
−
3
x
=
lim
x
→
+
∞
1
x
2
+
1
4
−
3
x
lim from{x toward +infinity} {{ sqrt{1+x^2} } over { 4x-3 } }= lim from{x toward +infinity} {{ {sqrt{1+x^2}} over x } over { {4x-3} over x } }= lim from{x toward +infinity} {{ sqrt{1 over x^2+1} } over { 4-3 over x } }
applichiamo ora i teoremi dell'algebra dei limiti:
lim
x
→
+
∞
1
+
x
2
4
x
−
3
=
lim
x
→
+
∞
lim
x
→
+
∞
1
x
2
+
1
4
−
lim
x
→
+
∞
3
x
=
1
4
lim from{x toward +infinity} {{ sqrt{1+x^2} } over { 4x-3 } }= lim from{x toward +infinity} {{ sqrt{lim from{x toward +infinity} {1 over x^2}+1} } over { 4-lim from{x toward +infinity} {3 over x} } }=1 over 4