lim(k f(x))=k lim(f(x))의 증명
{
k가 0일 경우는 trivial. k가 0이 아닌 것으로 가정. 극한값이 존재하고 L이라고 가정. 식
$\displaystyle 0<|x-c|<\delta\Rightarrow |f(x)-L|<\frac{\epsilon}k$
을 만족하는
$\displaystyle \delta$ 가 존재.
Someone is sure to complain that we put
$\displaystyle \epsilon/|k|$ rather than
$\displaystyle \epsilon$ at the end of the inequality above. Well, isn't
$\displaystyle \epsilon/|k|>0$ ? Yes. Doesn't the definition of limit require that for
any positive number there be a corresponding
$\displaystyle \delta?$ Yes.
Now, for
$\displaystyle \delta$ so determined, we assert that
$\displaystyle 0<|x-c|<\delta$ implies that
$\displaystyle |kf(x)-kL|=|k| |f(x)-L|<|k|\frac{\epsilon}{|k|}=\epsilon$
This shows that
$\displaystyle \lim_{x\to c}kf(x)=kL=k\lim_{x\to c}f(x)$
}
lim(f+g)=lim f + lim g의 증명
{
Let
$\displaystyle \lim_{x\to c}f(x)=L\textrm{ and }\lim_{x\to c}g(x)=M.$
다음을 만족하는
$\displaystyle \delta_1,\delta_2$ 가 존재.
$\displaystyle 0<|x-c|<\delta_1\Rightarrow|f(x)-L|<\frac{\epsilon}2$
$\displaystyle 0<|x-c|<\delta_2\Rightarrow|g(x)-M|<\frac{\epsilon}2$
Choose
$\displaystyle \delta=\min(\delta_1,\delta_2).$ Then
$\displaystyle 0<|x-c|<\delta$ implies that
$\displaystyle |f(x)+g(x)-(L+M)|$
$\displaystyle =|(f(x)-L)+(g(x)-M)|$
$\displaystyle \le|f(x)-L|+|g(x)-M|$
$\displaystyle <\frac{\epsilon}2+\frac{\epsilon}2=\epsilon$
첫번째 부등식은 삼각부등식. We've just shown that
$\displaystyle 0<|x-c|<\delta\Rightarrow |f(x)+g(x)-(L+M)|<\epsilon$
따라서
$\displaystyle \lim_{x\to c}(f(x)+g(x))=L+M=\lim_{x\to c}f(x)+\lim_{x\to c}g(x)$
}
lim(f-g)=lim f - lim g의 증명
{
$\displaystyle \lim_{x\to c}(f(x)-g(x))$
$\displaystyle =\lim_{x\to c}(f(x)+(-1)g(x))$
$\displaystyle =\lim_{x\to c}f(x)+\lim_{x\to c}(-1)g(x)$
$\displaystyle =\lim_{x\to c}f(x)+(-1)\lim_{x\to c}g(x)$
$\displaystyle =\lim_{x\to c}f(x)-\lim_{x\to c}g(x)$
}