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$\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? $\lim_{ x \to c }f(x)=L>\lim_{ x \to c }h(x)=M?$

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

No, limits don't preserve the strict inequality.

For example: $x^2 > x^4$ for all non-zero $x$ where $|x|<1$, yet they both have limit $0$ as $x \to 0$.

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  • $\begingroup$ The statement in the box is true ? $\endgroup$ – user402750 Mar 6 '17 at 15:31
  • $\begingroup$ Yes, that looks okay. $\endgroup$ – StackTD Mar 6 '17 at 15:32
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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 $f(x) > g(x)$

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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 $0.$

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