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Given $3$ functions $f,g,h:A \to \mathbb{R}$ and a cluster point $x_0$ for A such that $f(x)\leq g(x) \leq h(x)$, assume that $$\lim_{x \to x_0} f(x) = \lim_{x \to x_0} h(x)=L$$

By using ONLY THE DEFINITION OF LIMITS, prove that $\lim_{x \to x_0} g(x) = L$

Okay so I understand using squeeze theorem would prove this, but the question specifically asks for definition of limits, so how would I go about proving this? Just finding theoretical $\epsilon$ and $\delta$ and then plugging it into g(x)?

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    $\begingroup$ Use Heine's definition (sequential limit) and squeeze theorem. $\endgroup$ – Michael Galuza Aug 13 '15 at 15:16
  • $\begingroup$ could i use the definition of a limit to say $|f(x)-L| \leq |g(x)-L| \leq |h(x)-L| < \epsilon$ given $|x-x_0|< \delta$ and can conclude that way? $\endgroup$ – Lauren Bathers Aug 13 '15 at 15:22
  • $\begingroup$ en.wikipedia.org/wiki/Squeeze_theorem#Proof $\endgroup$ – Marco Cantarini Aug 13 '15 at 15:29
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Let $\varepsilon>0$. Choose $\delta>0$ such that $|x-x_0|<\delta$ implies $|f(x)-L|<\varepsilon$ and $\delta'>0$ such that $|x-x_0|<\delta'$ implies $|h(x)-L|<\varepsilon$. Then if $|x-x_0|<\min\{\delta,\delta'\}$, from $f(x)\leqslant g(x)\leqslant h(x)$ we have $$-\varepsilon<f(x)-L\leqslant g(x)-L\leqslant h(x)-L<\varepsilon, $$and hence $$|g(x)-L|<\varepsilon, $$ so that $$\lim_{x\to x_0}g(x)=L.$$

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  • $\begingroup$ thank you this explains it so well! $\endgroup$ – Lauren Bathers Aug 13 '15 at 15:27
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Given $\epsilon>0$, we can find $\delta_{1}$ such that for all $x \not =x_{0}$, $|x-x_{0}|<\delta_{1} \Rightarrow |f(x)-L|<0\epsilon$, and $\delta_{2}$ such that $|x-x_{0}|<\delta_{2} \Rightarrow |h(x)-L|<\epsilon$. Now, $h(x) \leq g(x) \leq f(x)$, so $h(x)-L \leq g(x) -L \leq f(x)-L$. Let $\delta=\min \lbrace {\delta_{1},\delta_{2}} \rbrace$. So if $|x-x_{0}|<\delta$, $|h(x)-L|<\epsilon$ so $-\epsilon <h(x)-L<\epsilon$ and $-\epsilon<f(x)-L<\epsilon$, hence $-\epsilon <g(x)-L < \epsilon$.

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It's really simple with Heine definition: function $f(x)$ have a limit $A$ at $x=x_0$ iff for any sequence $(x_n)$ such that $x_n\to x_0$ while $n\to\infty$ sequence $(f(x_n))$ tends to $A$.

So, let $(x_n)$ be any sequence such that $x_n\to x_0$, and $(y_n=f(x_n))\to L$ and $(z_n=g(x_n))\to L$. But $f(x_n)\le g(x_n)\le h(x_n)$ for any $x_n$. By squeeze theorem, $(g(x_n))\to L$, and it holds for every sequence $(x_n)$. Hence $\lim_{x\to x_0} g(x) = L$.

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Yes, you can proceed directly from $\epsilon,\delta$ formulations of the limit.

This is more or less exactly the statement of the squeeze theorem, so if you have seen a proof of that, try and remember it.

Otherwise, for a hint... enter image description here

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Let $\epsilon > 0$. Find a $\delta_f > 0$ so that if $|x - x_0| < \delta_f$, then $|f(x) - L| < \epsilon$. Similarly, find a $\delta_h > 0$ so that if $|x - x_0| < \delta_h$, then $|h(x) - L| < \epsilon$. Take $|x - x_0| < \min\{\delta_f, \delta_g\}$.

So, we have that $g(x) - L\le h(x) - L \le |h(x) - L| < \epsilon$

and $L - g(x) \le L - f(x) \le |L - f(x)| < \epsilon$

Thus $|g(x) - L| < \epsilon$.

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