2

No limit theorems allowed. Thanks!

We know it is something like

$$\left|\frac{1}{x_n}-\frac{1}{2}\right|=\left|\frac{2-x_n}{2x_n}\right|<\epsilon$$

Brian M. Scott
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Emily
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  • Yep, and now you can put an upper bound on the numerator and a lower bound on denominator, so that the fraction is bounded from above. – voligno Mar 19 '13 at 01:39

3 Answers3

6

Hint Since $x_n \to 2$, there exists some $N$ so that for all $n>N$ you have $x_n >1$. Then, for all $n>N$ you have:

$$\left|\frac{1}{x_n}-\frac{1}{2}\right|=\left|\frac{2-x_n}{2x_n}\right|<\left|\frac{2-x_n}{2}\right|$$

Brian M. Scott
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N. S.
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4

Let $\epsilon>0$. Since we care only about what happens when $\epsilon$ is small, we can also assume that $\epsilon<1$. (You’ll see in a bit why I want this.) Since $x_n\to 2$, there is an $m\in\Bbb N$ such that $|x_n-2|<\epsilon$ whenever $n\ge m$. Thus, for $n\ge m$ we have

$$\left|\frac1{x_n}-\frac12\right|=\left|\frac{2-x_n}{2x_n}\right|=\frac{|x_n-2|}{2|x_n|}<\frac{\epsilon}2\;,$$

because $|x_n-2|<\epsilon$ and $|x_n|>1$ (why?).

Brian M. Scott
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0

Continuing where you left off:

$$ |\frac{2-x_n}{2x_n}| = |\frac{2-x_n}{x_n - 2 + x_n - 2 + 4}| \leq \frac{|2-x_n|}{|x_n - 2| + |x_n - 2| + 4} $$

Since $x_n\to 2$, for any $\delta$ there exists an $N$ such that if $n>N, |2-x_n| < \delta$. Thus we need only find a $\delta(\epsilon)$ such that for a given $\epsilon > 0$,

$$ \frac{\delta}{2\delta + 4} \leq \epsilon $$

Hint: There's an easy simplification you can make to the above which makes finding the $\delta$ very simple.