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# 数学代写|实分析代写Real Analysis代考|One sided linlits

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## 数学代写|实分析代写Real Analysis代考|One sided linlits

There are cases where a function $f$ does not have a limit at a limit point $c$ of its domain $D$, but the restriction of the function $f$ to an interval at one side of $c$ (either right or left) may have a limit.

For example, the function $f: \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x)=\operatorname{sgn} x$ does not possess a limit at 0 but the restriction of $f$ to $(0, \infty)$ does have a limit at 0 and also the restriction of $f$ to $(-\infty, 0)$ does have a limit at 0 .
In:the former case we say that $f$ has a right hand limit at 0 and in the latter case we say that $f$ has a left hand limit at 0 .
Definitions.
Right hand limit. Let $D \subset \mathbb{R}$ and $f: D \rightarrow \mathbb{R}$ is a function. Let $c$ be a limit point of $D_1=D \cap(c, \infty)={x \in D: x>c}$.
$f$ is said to have a right hand limit $l(\in \mathbb{R})$ at $c$ if corresponding to a pre-assigned positive $\epsilon$ there exists a positive $\delta$ such that
$$|f(x)-l|<\epsilon \text { for all } x \in N^{\prime}(c, \delta) \cap D_1$$
i.e., $l-\epsilon<f(x)<l+\epsilon$ for all $x$ in $D$ satisfying $c<x<c+\delta$.
In this case we write $\lim _{x \rightarrow c+} f(x)=l$.
Left hand limit. Let $D \subset \mathbb{R}$ and $f: D \rightarrow \mathbb{R}$ is a function. Let $c$ be a limit point of $D_2=D \cap(-\infty, c)={x \in D: x<c}$.
$f$ is said to have a left hand limit $l(\in \mathbb{R})$ at $c$ if corresponding to a pre-assigned positive $\epsilon$ there exists a positive $\delta$ such that
$$|f(x)-l|<\epsilon \text { for all } x \in N^{\prime}(c, \delta) \cap D_2$$
i.e., $l-\epsilon<f(x)<l+\epsilon$ for all $x$ in $D$ satisfying $c-\delta<x<c$.
In this case we write $\lim _{x \rightarrow c-} f(x)=l$.

## 数学代写|实分析代写Real Analysis代考|Sequential criterion.

Let. $D \subset \mathbb{R}$ and $f: D \rightarrow \mathbb{R}$ be a function. Let $c$ be a limit point of $D_1=D \cap(c, \infty)$. Then $\lim _{x \rightarrow c+} f(x)=l$ if and only if for every sequence $\left{x_n\right}$ in $D_1$ converging to $c$, the sequence $\left{f\left(x_n\right}\right.$ converges to $l$.

Let $D \subset \mathbb{R}$ and $f: D \rightarrow \mathbb{R}$ be a function. Iet $c$ be a limit point of $D_2=D \cap(-\infty, c)$. Then $\lim _{x \rightarrow c-} f(x)=l$ if and only if for every sequence $\left{x_n\right}$ in $D_2$ converging to $c$, the sequence $\left{f\left(x_n\right}\right.$ corverges to $l$.

Note. It is possible that both the right hand limit and the left hand limit may exist, or both may not exist, or one of them exists while the other does not.
Worked Examples.

Let $f(x)=\operatorname{sgn} x$. Examine if $\lim {x \rightarrow 0^{+}} f(x)$ and $\lim {x \rightarrow 0^{-}} f(x)$ exist.
Here the domain $D$ of $f$ is $\mathbb{R}$.
Let $D_1=D \cap(0, \infty)$. Then $D_1={x \in \mathbb{R}: x>0}$. 0 is a limit point of $D_1 \cdot f(x)=1$ for all $x \in D_1$. Therefore $\lim _{x \rightarrow 0+} f(x)=1$.

Let $D_2=D \cap(-\infty, 0)$. Then $D_2={x \in \mathbb{R}: x<0}$. 0 is a limit point of $D_2 . f(x)=-1$ for all $x \in D_2$. Therefore $\lim _{x \rightarrow 0-} f(x)=-1$.

Note. Here both the right hand limit and the left hand limit of $f$ at 0 exist. $f$ is defined at 0 but $f(0) \neq \lim {x \rightarrow 0{+}} f(x)$ and also $f(0) \neq \lim _{x \rightarrow 0-} f(x)$.

## 数学代写|实分析代写Real Analysis代考|One sided linlits

$$|f(x)-l|<\epsilon \text { for all } x \in N^{\prime}(c, \delta) \cap D_1$$

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