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I've been working on this epsilon delta proof for the longest time now, and I can't quite get it.

Let $a>0$ and $b>0$. Use $\epsilon$, $\delta$ to prove that $\lim\limits_{x\to\ b} \frac{1}{a+x}$ = $\frac{1}{a+b}$.

I've found that I need to prove $|x-b|$ < $\epsilon$ $|a+x|$ $|a+b|$ and that I can do this by calling $\delta$ something greater than $|x-b|$, and that I also need to eliminate $|a+x|$ somehow, but that's where I'm struggling.

The concept of calling $\delta$ the minimum of a few values has been helpful, but I'm not sure how to apply it here.

Thanks for your thoughts.

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    $\begingroup$ if $|x - b| < 1$, then $|x| < |b| + 1$, so $|a + x| \le |a| + |x| < |a| + |b| + 1$. Now consider that $a, b$ are known constants. $\endgroup$ Commented May 27, 2016 at 22:26
  • $\begingroup$ Honestly: since $a,b>0$, we have that $f(b)$ is well defined; use continuity. $\endgroup$ Commented May 27, 2016 at 23:14

4 Answers 4

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$\forall \epsilon>0, \exists \delta>0$ such that $|x - b|< \delta \implies |\frac 1{a+x} - \frac 1{a+b}| < \epsilon$

$|\frac 1{a+x} - \frac 1{a+b}| = |\frac {b-x}{(a+x)(a+b)}| < \frac {\delta}{|(a+x)(a+b)|}$

If you can prove that $|(a+x)(a +b)|> 0$ then you are done. (which is pretty much what you told us, I just need to do certain things for myself)

$|a + x| = |a+b + (x-b)| \ge |a+b| - |x-b| \ge |a+b| - \delta$

let $\delta \le \min ( \frac {|a+b|}{2}, \epsilon M)$

In which case:

$|a + x| \ge \frac {|a+b|}{2}\\ |a + x||a+b| \ge \frac {(a + b)^2}2$

$|x-b|<\delta \le \min (\frac {(a + b)}2,\frac {(a + b)^2}2 \epsilon) \implies |\frac 1{a+x} - \frac 1{a+b}| < \epsilon$

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Let $\varepsilon > 0$. Want some $\delta > 0$ such that $0 < |x-b| < \delta$ implies $\left| \frac{1}{a + x} - \frac{1}{a + b} \right| < \varepsilon$. We start with the inequality $\left| \frac{1}{a + x} - \frac{1}{a + b} \right| < \varepsilon$ and try to manipulate it to give us insight into what $\delta$ should be. We have \begin{align} \frac{1}{a + x} - \frac{1}{a + b} &= \frac{a+b - (a+x)}{(a+x)(a+b)} \\ &= \frac{1}{b+a} \cdot \frac{b-x}{a+x} \\ \implies \left| \frac{1}{a + x} - \frac{1}{a + b} \right| &= \frac{1}{b+a} \cdot \frac{\left| b-x \right|}{\left| a+x \right|}. \tag{1} \end{align} We want to establish an upper bound for the LHS, which is accomplished by picking the largest possible RHS. This is found by maximizing $|b - x|$ and minimizing $|a - x|$. Our selection of $\delta$ determines which values of $x$ are allowed: $$ 0 < |x - b| < \delta \iff x \in (b-\delta, b) \cup (b, b+\delta). \tag{2} $$ Certainly $(2)$ implies the maximum of $|b - x|$ is $\delta$. The minimum of $a+x$ is $a + b - \delta$, and hence $$ \text{minimum of } |a + x| = \begin{cases} a + b - \delta & a + b - \delta > 0 \\ 0 & \text{otherwise} \end{cases}. \tag{3} $$ Armed with this info, let's try to pick a reasonable $\delta$ to bound the LHS of $(1)$. By $(3)$ we had better $$ \text{enforce $\quad \delta < a + b$}. \tag{4} $$ If we do, by $(1)$ we have \begin{align*} \left| \frac{1}{a + x} - \frac{1}{a + b} \right| & \leq \frac{1}{b+a} \cdot \frac{\delta}{a + b + \delta} \\ & \leq \frac{1}{b+a} \cdot \frac{\delta}{a + b}. \tag{5} \end{align*} Now, if we also $$ \text{enforce $\quad \delta < \varepsilon \cdot (a + b)^2$}, \tag{6} $$ then $(5)$ becomes $$ \left| \frac{1}{a + x} - \frac{1}{a + b} \right| \leq \frac{1}{b+a} \cdot \frac{\delta}{a + b} < \frac{1}{(a + b)^2} \cdot \varepsilon \cdot (a + b)^2 = \varepsilon. $$ That is, if we enforce $(4)$ and $(6)$, i.e., $\delta < \min\{a + b, \varepsilon \cdot (a+b)^2\}$, then $\left| \frac{1}{a + x} - \frac{1}{a + b} \right| < \varepsilon$, as desired.

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I think we should find a way to keep $|x-b|$ on the LHS of the inequality.

For a fixed small $\epsilon>0$, we want to find $\delta>0$ such that $|\frac{1}{a+x}-\frac{1}{a+b}|<\epsilon$ when $|x-b|<\delta$. Note that $|\frac{1}{a+x}-\frac{1}{a+b}|<\epsilon \Leftrightarrow \frac{|x-b|}{|a+x||a+b|}<\epsilon \Leftrightarrow \frac{|a+x||a+b|}{|x-b|}>\frac{1}{\epsilon}$.

Moreover, note that $|a+x|=|x-b+b+a|\leqslant |x-b|+|b+a|.$ Thus, it is sufficient to find $\delta>0$ such that $\frac{(|x-b|+|a+b|)|a+b|}{|x-b|}>\frac{1}{\epsilon}$, which is equivalent to $\frac{|a+b|^2}{|x-b|}>\frac{1}{\epsilon}-|a+b|$. And since $\epsilon>0$ is small, we can assume $\epsilon<\frac{1}{|a+b|}$. Then it is equivalent to $|x-b|<\frac{|a+b|^2}{\frac{1}{\epsilon}-|a+b|}$, thus it is sufficient to take $\delta=\epsilon|a+b|^2$ (since $\epsilon|a+b|^2<\frac{|a+b|^2}{\frac{1}{\epsilon}-|a+b|}$).

In $\epsilon-\delta$ language, we can state it as: for each fixed $0<\epsilon<\frac{1}{|a+b|}$, we have $|\frac{1}{a+x}-\frac{1}{a+b}|<\epsilon$ for all $x$ such that $|x-b|<\epsilon|a+b|^2$.

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The expression you want to keep small is $$ \frac{1}{a+x}-\frac{1}{a+b} =\frac{b-x}{(a+b)(a+x)} $$ and you can assume $x>0$ by taking $|x-b|<b$. Then $a+x>a$ and so $1/(a+x)<1/a$.

Therefore $$ \left|\frac{1}{a+x}-\frac{1}{a+b}\right|= \left|\frac{b-x}{(a+b)(a+x)}\right|< |b-x|\frac{1}{a(a+b)} $$ So, if $$ |b-x|<\varepsilon a(a+b) $$ you have the required inequality. So you can take $$ \delta=\min(b,\varepsilon a(a+b)) $$

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