As
x
≐
0
{\displaystyle \scriptstyle {x\doteq 0}}
y
{\displaystyle \scriptstyle {y}}
e
{\displaystyle \scriptstyle {e}}
x
≐
0
{\displaystyle \scriptstyle {x\doteq 0}}
y
{\displaystyle \scriptstyle {y}}
e
{\displaystyle \scriptstyle {e}}
As
x
≐
∞
{\displaystyle \scriptstyle {x\doteq \infty }}
y
{\displaystyle \scriptstyle {y}}
1
{\displaystyle \scriptstyle {1}}
≐
−
1
{\displaystyle \scriptstyle {\doteq -1}}
y
{\displaystyle \scriptstyle {y}}
In Chap. XVIII, Ex. 15, p. 233 , we will show how to calculate the value of
e
{\displaystyle \scriptstyle {e}}
Natural logarithms are those which have the number
e
{\displaystyle \scriptstyle {e}}
e
{\displaystyle \scriptstyle {e}}
log
e
v
{\displaystyle \scriptstyle {\log _{e}v}}
log
v
{\displaystyle \scriptstyle {\log v}}
Natural logarithms possess the following characteristic property: If
x
≐
0
{\displaystyle \scriptstyle {x\doteq 0}}
l
i
m
i
t
log
(
1
+
x
)
x
=
l
i
m
i
t
log
(
1
+
x
)
1
x
=
log
e
=
1.
{\displaystyle \scriptstyle {\operatorname {limit\;} {\frac {\log(1+x)}{x}}=\operatorname {limit\;} \log(1+x)^{\frac {1}{x}}=\log e=1.}}
24. Expressions assuming the form
∞
∞
{\displaystyle \scriptstyle {\frac {\infty }{\infty }}}
∞
{\displaystyle \scriptstyle {\infty }}
∞
÷
∞
{\displaystyle \scriptstyle {\infty \div \infty }}
Rule. Divide both numerator and denominator by the highest power of the variable occurring in either. Then substitute the value of the variable.
Illustrative Example 1. Evaluate
limit
x
=
∞
2
x
3
−
3
x
2
+
4
5
x
−
x
2
−
7
x
3
{\displaystyle \scriptscriptstyle {{\underset {x=\infty }{\operatorname {limit} }}\;{\frac {2x^{3}-3x^{2}+4}{5x-x^{2}-7x^{3}}}}}
Solution. Substituting directly, we get
limit
x
=
∞
2
x
3
−
3
x
2
+
4
5
x
−
x
2
−
7
x
3
=
∞
∞
{\displaystyle \scriptscriptstyle {{\underset {x=\infty }{\operatorname {limit} }}\;{\frac {2x^{3}-3x^{2}+4}{5x-x^{2}-7x^{3}}}={\frac {\infty }{\infty }}}}
x
3
{\displaystyle \scriptstyle {x^{3}}}
limit
x
=
∞
2
x
3
−
3
x
2
+
4
5
x
−
x
2
−
7
x
3
=
limit
x
=
∞
2
−
3
x
+
4
x
3
5
x
2
−
1
x
−
7
=
−
2
7
.
A
n
s
.
{\displaystyle \scriptscriptstyle {{\underset {x=\infty }{\operatorname {limit} }}\;{\frac {2x^{3}-3x^{2}+4}{5x-x^{2}-7x^{3}}}={\underset {x=\infty }{\operatorname {limit} }}\;{\frac {2-{\frac {3}{x}}+{\frac {4}{x^{3}}}}{{\frac {5}{x^{2}}}-{\frac {1}{x}}-7}}=-{\frac {2}{7}}.\quad Ans.}}
Prove the following:
1.
limit
x
=
∞
(
x
+
1
x
)
=
1
{\displaystyle \scriptscriptstyle {{\underset {x=\infty }{\operatorname {limit} }}\left({\frac {x+1}{x}}\right)=1}}
Proof.
limit
x
=
∞
(
x
+
1
x
)
=
limit
x
=
∞
(
1
+
1
x
)
=
limit
x
=
∞
(
1
)
+
limit
x
=
∞
(
1
x
)
=
1
+
0
=
1.
{\displaystyle {\begin{aligned}\scriptscriptstyle {{\underset {x=\infty }{\operatorname {limit} }}\left({\frac {x+1}{x}}\right)}&\scriptscriptstyle {={\underset {x=\infty }{\operatorname {limit} }}\left(1+{\frac {1}{x}}\right)}\\&\scriptscriptstyle {={\underset {x=\infty }{\operatorname {limit} }}(1)+{\underset {x=\infty }{\operatorname {limit} }}\left({\frac {1}{x}}\right)}\\&\scriptscriptstyle {=1+0=1.}\end{aligned}}}
Th. I, p. 18
