A constant can be moved through limit sign-any intuitive way to understand this?

[Juwane]

A constant can be moved through limit sign--any intuitive way to understand this?

We know that a constant can be moved through limit sign. However, according to my understanding, this result follows from the theorem that the limit of a product is equal to the product of the limits, and when one of the multiplicand of the product is a constant, then if we take the limit of the constant, it will equal to the constant itself.

But is there an intuitive or graphical way of showing that a constant can be moved through a limit sign?

 

[l'Hôpital]

Suppose we were considering the following:
<br /> lim_{x \rightarrow a} \ c f(x)<br />
Where c is a constant.

Then, we can agree that
<br /> c f(x) = \underbrace{f(x) + f(x) + ... + f(x)}_{c}<br />

Thus,
<br /> lim_{x \rightarrow a} \ c f(x) = lim_{x \rightarrow a} \underbrace{f(x) + f(x) + ... + f(x)}_{c}<br />

Would you be willing to believe the sum of the limits is the limit of the sum? If so, we're done.

 

[HallsofIvy]

l'Hôpital, are you asserting that this is only true if c is a positive integer? That's the only case in which
c f(x) = \underbrace{f(x) + f(x) + ... + f(x)}_{c}

(Added- Ah, I see, Juwant asked for an "intuitive" way of seeing it.)

From the definition of "limit":
If \lim_{x\to a} cf(x)= L, c a (non zero) constant, then given any \epsilon&gt; 0, there exist \delta&gt; 0 such that if |x-a|&lt;\delta, then |cf(x)- L|&lt; \epsilon. Therefore, given any \epsilon&gt; 0, |c|\epsilon is also greater than 0 and there exist \delta&gt; 0 such that if |x- a|&lt; \delta, |cf(x)- L|&lt; |c|\epsilon. Now, |c||f(x)- L/c|&lt; |c|\epsilon so |f(x)- L/c|&lt; \epsilon so \lim_{x\to a} f(x)= L/c.

That is, if \lim_{x\to a} cf(x)= L, then \lim_{x\to a}f(x)= L/c which is the same as c \lim_{x\to a}f(x)= L.

If c= 0, then cf(x)= 0 for all x so \lim_{x\to a} cf(x)= 0= 0(\lim_{x\to a} f(x)) as well.