这是一份anu澳大利亚国立大学MATH2222/MATH6222的成功案例

For a function $f, \lim {x \rightarrow a} f(x)=L$ if and only if $$ \lim {x \rightarrow a^{+}} f(x)=\lim {x \rightarrow a^{-}} f(x)=L . $$ Theorem $5.3 .2 \mathrm{can}$ be argued in an interesting way from the logical structure of the definitions of the limits involved, thus we’ll provide a rather conversational argument. In arguing the $\Rightarrow$ direction, we would pick an $\epsilon>0$ and note that there exists $\delta>0$ such that $x \in D N{\delta}(a)$ implies $|f(x)-L|<\epsilon$. However, the inequality $x \in D N_{\delta}(a)$ is weaker than either $a<x<a+\delta$ or $a-\delta<x<a$. Thus the statement
$$
x \in D N_{\delta}(a) \Rightarrow|f(x)-L|<\epsilon
$$
is stronger than either
$$
a<x<a+\delta \Rightarrow|f(x)-L|<\epsilon
$$
or
$$
a-\delta<x<a \Rightarrow|f(x)-L|<\epsilon .
$$
Thus the existence of $\lim {x \rightarrow a} f(x)$ implies the existence of $\lim {x \rightarrow a^{+}} f(x)$ and $\lim _{x \rightarrow a^{-}} f(x) .$

MATH2222/MATH6222 COURSE NOTES ：

$$
1+y+y^{2}+\cdots+y^{n}=\frac{1-y^{n+1}}{1-y}
$$
and
$$
1-y+y^{2}-\cdots+(-y)^{n}=\frac{1-(-y)^{n+1}}{1+y}
$$
If we know that $0\frac{1}{1+y}
$$
Since $0<\epsilon<1 / a$, we have $0<\epsilon a<1$, so that Eqs. (5.37) and (5.38) both hold for $y=\epsilon a$. Thus
$$
\frac{1}{1+\epsilon a}<1-\epsilon a(1-\epsilon a) \quad \text { and } \quad 1+\epsilon a<\frac{1}{1-\epsilon a}
$$