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If
lim
x
→
0
(
tan
(
π
4
+
x
)
)
1
x
−
e
2
x
2
\lim_{x \rightarrow 0} \frac{(\tan(\frac{\pi}{4} + x))^{\frac{1}{x}} - e^2}{x^2}
lim
x
→
0
x
2
(
t
a
n
(
4
π
+
x
)
)
x
1
−
e
2
is…
Mathematics
Problem
If
lim
x
→
0
(
tan
(
π
4
+
x
)
)
1
x
−
e
2
x
2
\lim_{x \rightarrow 0} \frac{(\tan(\frac{\pi}{4} + x))^{\frac{1}{x}} - e^2}{x^2}
lim
x
→
0
x
2
(
t
a
n
(
4
π
+
x
)
)
x
1
−
e
2
is equal to
(
p
q
)
e
2
(\frac{p}{q})e^2
(
q
p
)
e
2
,
(
p
,
q
∈
N
)
(p, q \in \mathbb{N})
(
p
,
q
∈
N
)
, then the minimum value of
(
p
+
q
)
(p + q)
(
p
+
q
)
is (a)
5
5
5
(b)
6
6
6
(c)
7
7
7
(d)
9
9
9
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