# Any proof that verify why the limit of the difference is the difference of the limits?

I did a research on internet and books about why the difference of the limits is the difference of the limits, but i didn't get any result of this proof. I would appreciate if somebody can help me. Thanks. :)

-
You're asking about the algebraic properties of limits. Here are a few good proofs of some other algebraic properties of limits, including the one you're asking about: tutorial.math.lamar.edu/Classes/CalcI/LimitProofs.aspx –  danielson Aug 24 '14 at 3:25
If you can prove the limit of the sum is the sum of the limits. Then your question should follow easily by adding a negative. –  TylerHG Aug 24 '14 at 3:29
You can prove it in the almost the exact same way as the sum since $|f(x) - g(x) - (K - L)| \leq |f(x) -K| + |g(x) - L|$ just as in the proof for the sum (using the notation on the site). –  danielson Aug 24 '14 at 3:34
Yes it very acceptable! In fact I do not think that most mathematicians would even bother proving it separately as it is essentially the same thing as danielson said. Mathematicians are lazy ;) –  TylerHG Aug 24 '14 at 3:37
Like danielson shows, $|f(x) - g(x) - (K-L)| \leq |f(x) - K| + |g(x) - L|$. This inequality comes from triangle inequality. –  Gudson Chou Aug 24 '14 at 4:06

Consider $f(x)$ and $g(x)$ are two functions on real domain whose limits exist for $x=a$.

Given: $\lim_{x \to a} f(x)$ = $K$ and $\lim_{x \to a} g(x)$ = $L$

Let $\varepsilon$ > 0 then there exist $\alpha$ > $0$ and $\beta$ > $0$ such that ,

$|f(x)-K |$< $\varepsilon$/2 ,whenever $0$<|$x$-$a$|<$\alpha$ ,and $|g(x)-L |$< $\varepsilon$/2 , whenever $0$<|$x$-$a$|<$\beta$

Choose $\gamma$= $min${$\alpha$,$\beta$}

Now we need to show that $|f(x)+g(x)-(K+L)|$

Assume that we have $0$<$|x-a|$<$\gamma$. Then we have,

$|f(x)+g(x)-(K+L)|$=$|(f(x)-K)+(g(x)-L)|$ < $|(f(x)-K)|$ + $|(g(x)-L)|$ < $\varepsilon$

In our third step we used the fact that, by our choice of $\gamma$, we also have $0$<|$x$-$a$|

So we can use initial statements in our proof.

Now replace $g(x)$ by $(-1)g(x)$ and you will get your proof.

-

if $\mathfrak{a} = \{a_n\}_{n=1,...}$ is a sequence of (let us say) real numbers, and $a$ is a real number, then we may define the statement $L(\mathfrak{a},a)$ as meaning $\forall \epsilon > 0 .\exists N_{\epsilon}.n \gt N_{\epsilon} \Rightarrow |a_n - a| \lt \epsilon$.

so you wish to show that $$L(\mathfrak{a},a) \land L(\mathfrak{b},b) \Rightarrow L(\mathfrak{a}-\mathfrak{b},a-b)$$ where, by definition, $\mathfrak{a}-\mathfrak{b} = \{a_n-b_n\}_{n=1,...}$

for the statement $L(\mathfrak{a}-\mathfrak{b},a-b)$ to be true then given any $\epsilon \gt 0$ we must be able to find $M_{\epsilon}$ such that $$m \gt M_{\epsilon} \Rightarrow |(a_m - b_m) - (a-b)| \lt \epsilon$$ the condition may be written: $$|(a_m - a) - (b_m-b)| \lt \epsilon$$ and from the triangle equality we know that: $$|(a_m - a) - (b_m-b)| \le |(a_m - a)| + |(b_m-b)|$$ since $L(\mathfrak{a},a)$ and $L(\mathfrak{b},b)$ we can find integers $J_{\frac12\epsilon}$ and $K_{\frac12\epsilon}$ such that $$m \gt J_{\frac12\epsilon} \rightarrow |(a_m - a)| \lt \frac12\epsilon \\ m \gt K_{\frac12\epsilon} \rightarrow |(b_m - b)| \lt \frac12\epsilon$$ so choose an integer $M \gt \max( J_{\frac12\epsilon}, K_{\frac12\epsilon})$ now if $m \gt M$ $$|(a_m - a) - (b_m-b)| \le |(a_m - a)| + |(b_m-b)| \le \frac12\epsilon+ \frac12\epsilon = \epsilon$$

-
i think the difficulty we sometimes have with these arguments arises because the technical jiggery-pokery with absolute values & triangle inequalities tends to obscure the underlying logic of making statements about infinite processes. however this awkwardness shouldn't prevent us from admiring the lucidity of what is now known as "Cauchy's criterion" for convergence - as you probably know, this has the far-reaching consequence of allowing the reals to be constructed as a "topological" completion of the rationals. without completeness a "convergent" sequence is not guaranteed to have a limit –  David Holden Aug 24 '14 at 4:15