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I don't know if this property is right:

If $f \leq g \leq h$, $\forall x\in\mathbb{R}$ and $\lim_{x \to a} f = L$, $\lim_{x \to a} h = L$, then $\lim_{x \to a} g = L$.

My intention on these question is a generalization, I mean, let $$f, g, h: A\subseteq\mathbb{R^n} \longrightarrow \mathbb{R}$$ such that $$f\leq g\leq h, \forall \bar{x}\in A$$ and $$\lim_{\bar{x} \to \bar{x_0}} f = \lim_{\bar{x} \to \bar{x_0}} h = L$$ then $$\lim_{\bar{x} \to \bar{x_0}} g = L$$

Blue Tomato
  • 789

2 Answers2

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This is known as the squeeze theorem. It follows by $$ L = \lim_{x\to x_0} f(x) \leq \liminf_{x\to x_0} g(x)\leq \limsup_ {x\to x_0} g(x) \leq \lim_ {x\to x_0} h(x)=L $$.

Edit: add missing $\leq$

BAND
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Let $(X,d_{X})$ be a metric space and $f:X\to\mathbb{R}$, $g:X\to\mathbb{R}$ and $h:X\to\mathbb{R}$ be functions.

Moreover, let $E\subseteq X$ such that $a\in X$ is an adherent point of $E$ and $L\in\mathbb{R}$.

Also, let us assume that $f$ and $h$ converges to $L$ as $x$ approaches $a$ along $E$ s.t. $f\leq g\leq h$ for every $x\in X$.

Then for every $\varepsilon > 0$, there corresponds $\delta^{1}_{\varepsilon} > 0$ s.t. for every $x\in E$ one has that \begin{align*} d_{X}(x,a) < \delta^{1}_{\varepsilon} \Rightarrow |f(x) - L| < \varepsilon \Rightarrow L - \varepsilon < f(x) < L + \varepsilon \end{align*}

Similarly, for the same $\varepsilon > 0$, there corresponds a $\delta^{2}_{\varepsilon} > 0$ s.t. for every $x\in E$ one has that \begin{align*} d_{X}(x,a) < \delta^{2}_{\varepsilon} \Rightarrow |h(x) - L| < \varepsilon \Rightarrow L - \varepsilon < h(x) < L + \varepsilon \end{align*}

Hence for every $\varepsilon > 0$, there corresponds a $\delta = \min\{\delta^{1}_{\varepsilon},\delta^{2}_{\varepsilon}\}$ s.t. for every $x\in E$ we have that \begin{align*} d_{X}(x,a) < \delta \Rightarrow L - \varepsilon < f(x) \leq g(x) \leq h(x) < L + \varepsilon \Rightarrow |g(x) - L| < \varepsilon \end{align*}

and we are done.

Hopefully this helps!

user0102
  • 21,572
  • Man! I have one little bit interesting question. We know that norms implies metric, so is it a little bit more general to prove this property on normed spaces? – Blue Tomato Mar 21 '21 at 13:47
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    @Someone As long as you have a metric space, this proposition holds. Therefore you can substitute the metric by the norm of the difference, which is essentially the same thing. – user0102 Mar 21 '21 at 19:24