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Hwangjy9/고등학교의 특별한 극한값
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User:Hwangjy9
함수의 극한값
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lim
x
→
0
sin
x
x
=
1
{\displaystyle \lim _{x\to 0}{\frac {\sin x}{x}}=1}
lim
x
→
0
ln
(
1
+
x
)
x
=
1
{\displaystyle \lim _{x\to 0}{\frac {\ln(1+x)}{x}}=1}
lim
x
→
0
e
x
−
1
x
=
1
{\displaystyle \lim _{x\to 0}{\frac {e^{x}-1}{x}}=1}
증명
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11
11
11