Prove that $\lim cf(x) = c\lim f(x)$ Well, I'm having trouble with these types of proofs, based on épsilon delta. I've found these proofs: http://en.wikibooks.org/wiki/Calculus/Proofs_of_Some_Basic_Limit_Rules but I cannot understand. Can somebody explain the scalar product limit proof intuitively?
 thank you so much
 A: Here is how I explain the $\epsilon$-$\delta$ limit definition to my students:
We wish to explain the meaning of the statement $\displaystyle\lim_{x\to c}f(x)=L$.  Suppose you are trying to force the function $f$ to take a value close to $L$.  That is, someone tells you that you must get $f(x)$ to be close to $L$, with only a little bit of error allowed (this "little bit of error" is the $\epsilon$).  Then if it is really true that $\displaystyle\lim_{x\to c}f(x)=L$, then you should be able to force $f(x)$ to be close to $L$ by first forcing $x$ to be close to $c$.  The question is how close does $x$ need to be to $c$?  That distance will be your $\delta$.
That is, if you can force $x$ to be within $\delta$ of $c$, that will force $f(x)$ to be within your allowed error of $\epsilon$ of $L$.
Thus the statement "$\displaystyle\lim_{x\to c}f(x)=L$" may be translated as "We can force $f(x)$ to be as close as we like to $L$ by merely forcing $x$ to be sufficiently close to $c$.
Now an intuitive proof of the property you mentioned.  Suppose $\displaystyle\lim_{x\to c}f(x)$ holds.  Let us first show that $\displaystyle\lim_{x\to c}5f(x)=5\lim_{x\to c}f(x)=5L$.  That is, I want to show that I can force $5f(x)$ to be as close to $5L$ as I like just by forcing $x$ to be sufficiently close to $c$.  Well suppose that I want $5f(x)$ to be within $\dfrac{1}{100}$ of $5L$.  I just make $x$ be close enough to $c$ to force $f(x)$ to be within $\dfrac{1}{500}$ of $L$.  That is, so that $|f(x)-L|<\frac{1}{500}$.  Now just multiply both sides of that inequality by $5$.  We get
$$
|5f(x)-5L|<\dfrac{5}{500}=\dfrac{1}{100}.
$$
The very inequality we wanted!  Now replace the $5$ with any $a$ and the $\frac{1}{100}$ by any $\epsilon>0$, and we are done.  (Note: when I said "make $x$ close enough to $c$", that "close enough" would be the $\delta$.)
