# if $f(x)>g(x)$ can write? $\lim_{ x \to c }f(x)=L>\lim_{ x \to c }h(x)=M?$

$\text{Let }A,B⊆\mathbb{R} \text{ suppose that } c \in \mathbb{R}\text{is a }\text{cluster point or accumulation point of} A, B$

$\text{and } f:A\to \mathbb{R},g:B\to \mathbb{R} \text{be two real valued function deﬁned on A,B}, \text{suppose that} U \text{is a deleted neighborhood of c } \text{such that} : \\f(x)≥g(x) \text{for every }x\in U∩A∩B \\\text{and }\lim_{ x \to c }f(x)=L\in\mathbb{R},\lim_{ x \to c }h(x)=M\in\mathbb{R}\\\text{then }\lim_{ x \to c }f(x)=L≥\lim_{ x \to c }h(x)=M$

is it right ?

now : if $f(x)>g(x)$ can write? $\lim_{ x \to c }f(x)=L>\lim_{ x \to c }h(x)=M?$

now : if $f(x)>g(x)$ can write? $\lim_{ x \to c }f(x)=L>\lim_{ x \to c }h(x)=M?$
For example: $x^2 > x^4$ for all non-zero $x$ where $|x|<1$, yet they both have limit $0$ as $x \to 0$.
If you mean $h(x) = g(x)$ then it's not true... For example $A,B = \mathbb{R}\backslash{c} , f(x) = e^{|x-c|}, g(x)=e^{\frac{|x-c|}2}$ both lim at $x=c$ is 1 and yet for every other point $f(x) > g(x)$
The statement in the box is OK. However consider $f(x)=|x|,\ h(x)=x^2$ on a deleted neighborhood of $0.$ Then $L=M=0$ but $f(x)>h(x)$ near $0.$