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Putnam
2021 Putnam
A2
Putnam 2021 A2
Putnam 2021 A2
Source:
December 5, 2021
Putnam
Putnam 2021
Problem Statement
For every positive real number
x
x
x
, let
g
(
x
)
=
lim
r
→
0
(
(
x
+
1
)
r
+
1
−
x
r
+
1
)
1
r
.
g(x)=\lim_{r\to 0} ((x+1)^{r+1}-x^{r+1})^{\frac{1}{r}}.
g
(
x
)
=
r
→
0
lim
((
x
+
1
)
r
+
1
−
x
r
+
1
)
r
1
.
Find
lim
x
→
∞
g
(
x
)
x
\lim_{x\to \infty}\frac{g(x)}{x}
lim
x
→
∞
x
g
(
x
)
. By the Binomial Theorem one obtains\\
lim
x
→
∞
lim
r
→
0
(
(
1
+
r
)
+
(
1
+
r
)
r
2
⋅
x
−
1
+
(
1
+
r
)
r
(
r
−
1
)
6
⋅
x
−
2
+
…
)
1
r
\lim_{x \to \infty} \lim_{r \to 0} \left((1+r)+\frac{(1+r)r}{2}\cdot x^{-1}+\frac{(1+r)r(r-1)}{6} \cdot x^{-2}+\dots \right)^{\frac{1}{r}}
lim
x
→
∞
lim
r
→
0
(
(
1
+
r
)
+
2
(
1
+
r
)
r
⋅
x
−
1
+
6
(
1
+
r
)
r
(
r
−
1
)
⋅
x
−
2
+
…
)
r
1
\\
=
lim
r
→
0
(
1
+
r
)
1
r
=
e
=\lim_{r \to 0}(1+r)^{\frac{1}{r}}=\boxed{e}
=
lim
r
→
0
(
1
+
r
)
r
1
=
e
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