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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.) $b_{n} > d_{n} > 0$.

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Hint: Follow your proof if $d_n$ and $b_n$ are negative. – Thomas Andrews Feb 28 '12 at 14:08
I need them to be positive ... if I prove this, I might have a prove for the following post… – rtybase Feb 28 '12 at 14:33
up vote 3 down vote accepted

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.

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Yep ... good one. – rtybase Feb 28 '12 at 15:18
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: $$\alpha \cdot (b_{n}-d_{n}) - \varepsilon \cdot (b_{n}+d_{n}) < a_{n} - c_{n} < \alpha \cdot (b_{n}-d_{n}) + \varepsilon \cdot (b_{n}+d_{n})$$ – rtybase Feb 28 '12 at 15:57
@rtybase: Seems very sensible, informally that restriction should be enough. – André Nicolas Feb 28 '12 at 16:06

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