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If $$\liminf_{x\to \infty}f(x)>M,$$ for some $M>0$, where $f$ is continuous function on $\mathbb R$.

Does this imply that: there exist $x_{o}$ such that for all $x\geq x_{o}$ we have $f(x)>M$? If so, how I can prove it?

I also have another question, just to be sure: If $f(x)\leq g(x)$ for all $x\in \mathbb R$, both are continuous on $\mathbb R$, then $\liminf_{x\to\infty}f(x)\leq \liminf_{x\to\infty} g(x)$

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    Write down the definition of $\liminf$. Then try to assume that what you wrote is false. That is there is a sequence of $x_n$ going to infinity with $f(x_n)<=M$.2012-04-02
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    I know that the first argument is true for sequences: If $\liminf_{n\to \infty}x_{n}>M$ then there exists $n_{o}$ such that $x_{n}>M$ for all $n\geq n_{o}$. But is it true also for functions!?2012-04-02
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    Using Alex R.'s argument, a contradiction will happen. However, I'm going to do this question in another way for fun.2016-02-11

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  1. Let $L = \liminf_{x\to\infty}f(x) = \sup_{N>0}\inf_{x\ge N}f(x)$. Then we have $L>M$, so $M$ isn't a supremum of $\left\{\inf_{x\ge N}f(x)\mid N>0\right\}$. Therefore, $\exists x_0>0$ such that $\inf_{x\ge x_0}f(x)>M$. Hence, $f(x) > M \;\forall x \ge x_0$.
  2. The answer is yes. We're given $$f(x)\leq g(x) \;\forall x \in \Bbb R.\tag{1}$$ Let $N>0$. We first take infimum on the LHS over all $x\ge N$. $$\inf_{x\ge N}f(x) \le g(x) \;\forall x \ge N\tag{2}$$ Then $\inf_{x\ge N}f(x)$ is a lower bound of $\left\{g(x)\mid x \ge N\right\}$, while $\inf_{x\ge N}g(x)$ is the greatest lower bound of $\left\{g(x)\mid x \ge N\right\}$. Therefore, one has $$\inf_{x\ge N}f(x) \le \inf_{x\ge N}g(x).\tag{3}$$ From (1) to (3), we realize that we can take infimum on both sides. Similarly, we can take supremum over $N>0$ on both sides. $$\sup_{N>0}\inf_{x\ge N}f(x) \le \sup_{N>0}\inf_{x\ge N}g(x)\tag{4}$$ In other words, we have the statement in the OP. $$\liminf_{x\to\infty}f(x) \le \liminf_{x\to\infty}g(x)$$