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Let $g$ be continuous on the interval $[a, b]$, and a $< c < b$ with $g_{(c)}<0$

Prove, using the epsilon and delta characterisation of a limit, that $\lim\limits_{x \to {c}}\frac{1}{g{(x)}} = \frac{1}{g{(c)}}$ Hence deduce that $\frac{1}{g}$ is continuous at $c$

I know that there exists $g {(x}$) smaller than $\frac{g {(c)}}{2} < 0$ and I can use this information somehow but other than that I am not sure how to approach this question. Is there any special laws I need to follow?

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  • $\begingroup$ Start by the very definition: for every $\varepsilon > 0$ there exists a $\delta > 0$ such that... and then use that $g$ is continuous. $\endgroup$ – Mike Nov 19 '15 at 17:15
  • $\begingroup$ yeah i know that much.. its continuing from there $\endgroup$ – Jodie Dougal Nov 19 '15 at 17:22
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Since $g$ is continuous on $[a,b]$, then for all $\epsilon>0$, and for all $c\in [a,b]$, there exists a $\delta(\epsilon,c)>0$ such that

$$|g(x)-g(c)|<\left(\frac{g^2(c)}{2}\right)\epsilon \tag 1$$

whenever, $|x-c|<\delta(\epsilon,c)$.

So, if we take $\epsilon=\frac12|g(c)|$, then we see that $\frac32g(c)<g(x)<\frac12g(c)$, since $g(c)<0$, whenever $|x-c|<\delta(g(c),c)$.

Therefore, we can write for all $\epsilon>0$

$$\begin{align} \left|\frac{1}{g(x)}-\frac{1}{g(c)}\right|&=\frac{|g(x)-g(c)|}{|g(x)|\,|g(c)|}\\\\ &\le\frac{|g(x)-g(c)|}{\frac12|g(c)|\,|g(c)|}\\\\ &<\epsilon \end{align}$$

whenever $|x-c|<\min\left(\delta(\epsilon,c),\delta(g(c),c)\right)$

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  • $\begingroup$ You're welcome. My pleasure. $\endgroup$ – Mark Viola Nov 19 '15 at 18:01
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As I wrote in my comment, let us start by what we have to show: for all $\varepsilon > 0$ there exists a $\delta > 0$ such that \begin{equation} \Bigg| \frac{1}{g(x)} - \frac{1}{g(c)} \Bigg| = \Bigg| \frac{g(c) -g(x)}{g(x) g(c)} \Bigg| = \frac{\lvert g(c) - g(x) \rvert}{ \lvert g(x) \rvert \lvert g(c) \rvert} < \varepsilon \end{equation} for all $x \in [a,b]$ with $\lvert x - c \rvert < \delta$.

Fix $\varepsilon > 0$. By continuity of $g$, there exists a $\delta_1 >0 $ such that \begin{equation} \lvert g(c) -g(x) \rvert < \frac{\lvert g(c) \rvert^2}{2} \varepsilon \end{equation} for all $x \in [a,b]$ with $\lvert x-c \rvert < \delta_1$. Moreover, since $g(c) < 0$, we find a $\delta_2 > 0$ such that \begin{equation} g(x) \leq \frac{g(c)}{2} < 0 \end{equation} for all $x \in [a,b]$ with $\lvert x- c \rvert < \delta_2$. Note that for those $x$, $\lvert g(x) \rvert \geq \frac{ \lvert g(c) \rvert}{2} $.

Let us now choose a $\delta \in \big(0, \min \{ \delta_1, \delta_2 \} \big)$. Then \begin{equation} \Bigg| \frac{1}{g(x)} - \frac{1}{g(c)} \Bigg| \leq \frac{2}{ \lvert g(c) \rvert^2} \lvert g(c) - g(x) \rvert < \varepsilon \end{equation} for all $x \in [a,b]$ with $\lvert x-c \rvert < \delta$.

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  • $\begingroup$ Just like mine. ;-)) $\endgroup$ – Mark Viola Nov 19 '15 at 17:47

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