In arithmetic there is a property that if $\frac{a}{b}=\frac{c}{d}=\alpha$ then $\frac{a-c}{b-d}=\frac{a+c}{b+d}=\alpha$, with the first we assume $b\neq d$.

With the limits, for example, if $\lim_{n\to\infty }\frac{a_{n}}{b_{n}}=\lim_{n\to\infty }\frac{c_{n}}{d_{n}}=\alpha$, assuming $b_{n}> 0$ and $d_{n}> 0$, then it is easy to prove that $\lim_{n\to\infty }\frac{a_{n}+c_{n}}{b_{n}+d_{n}}=\alpha$. I.e.

$$\alpha -\varepsilon < \frac{a_{n}}{b_{n}} < \alpha +\varepsilon$$ $$\alpha -\varepsilon < \frac{c_{n}}{d_{n}} < \alpha +\varepsilon$$

From which

$$\left ( \alpha -\varepsilon \right )\cdot b_{n} < a_{n} < \left ( \alpha +\varepsilon \right )\cdot b_{n}$$ $$\left ( \alpha -\varepsilon \right )\cdot d_{n} < c_{n} < \left ( \alpha +\varepsilon \right )\cdot d_{n}$$

And as a result $$\left ( \alpha -\varepsilon \right )\cdot \left ( b_{n} + d_{n} \right ) < a_{n}+c_{n} < \left ( \alpha +\varepsilon \right )\cdot \left ( b_{n} + d_{n} \right )$$

How about $\lim_{n\to\infty }\frac{a_{n}-c_{n}}{b_{n}-d_{n}}$ ?

Let's ignore the case when $\left \{ a_{n} \right \}$, $\left \{ b_{n} \right \}$, $\left \{ c_{n} \right \}$ and $\left \{ d_{n} \right \}$ converge, this is easy to prove. Stick with $\lim_{n\to\infty }\frac{a_{n}}{b_{n}}=\lim_{n\to\infty }\frac{c_{n}}{d_{n}}=\alpha$ though and (e.g.) $b_{n} > d_{n} > 0$.

  • $\begingroup$ Hint: Follow your proof if $d_n$ and $b_n$ are negative. $\endgroup$ – Thomas Andrews Feb 28 '12 at 14:08
  • $\begingroup$ I need them to be positive ... if I prove this, I might have a prove for the following post math.stackexchange.com/questions/72942/… $\endgroup$ – rtybase Feb 28 '12 at 14:33

Let $a_n=n+\frac{47}{n}$ and let $b_n=n+\frac{1}{n}$. Let $c_n=d_n=n$.

Then $$\lim_{n\to\infty}\frac{a_n}{b_n}=\lim_{n\to\infty}\frac{c_n}{d_n}=1.$$ However, $$\lim_{n\to\infty}\frac{a_n-c_n}{b_n-d_n}=47.$$

Remark: We can in a similar way produce any desired kind of behaviour: "$0/0$" is indeterminate. Or, from a numerical analysis point of view, subtraction can lead to a catastrophic loss of precision.

The issue cannot be resolved by asking that $b_n-d_n$ not approach $0$, for we can multiply $a_n$, $b_n$, $c_n$ and $d_n$ by $n$.

One could use the simpler example $a_n=n+17$, $b_n=n+1$, $c_n=d_n=n$. But the one given in the main post was the first one I thought of, so it seemed reasonable to preserve the "$0/0$" intuition behind the example.

  • $\begingroup$ Yep ... good one. $\endgroup$ – rtybase Feb 28 '12 at 15:18
  • $\begingroup$ In order to address this, I think, there should be one extra condition that $\lim_{n\to\infty }\frac{b_{n}}{d_{n}}$ exists and it is $> 1$. This will help to address the inequality: $$\alpha \cdot (b_{n}-d_{n}) - \varepsilon \cdot (b_{n}+d_{n}) < a_{n} - c_{n} < \alpha \cdot (b_{n}-d_{n}) + \varepsilon \cdot (b_{n}+d_{n})$$ $\endgroup$ – rtybase Feb 28 '12 at 15:57
  • $\begingroup$ @rtybase: Seems very sensible, informally that restriction should be enough. $\endgroup$ – André Nicolas Feb 28 '12 at 16:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.