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:
$$lim_{x \rightarrow a} \ c f(x)$$
Where c is a constant.

Then, we can agree that
$$c f(x) = \underbrace{f(x) + f(x) + ... + f(x)}_{c}$$

Thus,
$$lim_{x \rightarrow a} \ c f(x) = lim_{x \rightarrow a} \underbrace{f(x) + f(x) + ... + f(x)}_{c}$$

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 $\epsilon> 0$, there exist $\delta> 0$ such that if $|x-a|<\delta$, then $|cf(x)- L|< \epsilon$. Therefore, given any $\epsilon> 0$, $|c|\epsilon$ is also greater than 0 and there exist $\delta> 0$ such that if $|x- a|< \delta$, $|cf(x)- L|< |c|\epsilon$. Now, $|c||f(x)- L/c|< |c|\epsilon$ so $|f(x)- L/c|< \epsilon$ 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.