Question

Question 6 of the Unit 1 exam supposes f satisfies:

f(x+y) = f(x) + f(y) + x^2y + xy^2

It also supposes (A) the limit of f(x)/x as x nears 0 is 1. It says to evaluate f(0).

The solutions says that if A, then (B) the limit of f(x) as x nears 0 is 0. I see that given B, f(0) = 0, but how does A get you to B?

It seems more obvious that since

f(0+0) = f(0) + f(0) + 0^2*0 + 0*0^2,

it has to be that f(0) = f(0) + f(0),

but then f(0) = 0.

It is clear how to go on from here, can someone explain the connection between A and B?

Answer 1

Ah, I see....

Given lim [f(x)/x] (x->0) = 1:

Then: ln(lim (x->0) [f(x)/x]) = 0.
Then: lim (x->0) [ln f(x) - ln x] = 0
Then: lim (x->0) [ln f(x)] = lim (x->0) [ln x]
Then: ln [lim (x->0) f(x)] = ln [lim (x->0) x]
Then: e^(ln [lim (x->0) f(x)]) = e^(ln [lim (x->0) x])
Then: lim (x->0) f(x) = lim (x->0) x
Since lim (x->0) x = 0, so lim (x->0) f(x) = 0.

Needs most of the tricks....

Answer 2

yes, your way works.
another way is start with
\[ \lim_{x\rightarrow 0} \frac{f(x)}{x} = 1 \]
being pedantic we can also take the limit of the right side
\[ \lim_{x\rightarrow 0} \frac{f(x)}{x} = \lim_{x\rightarrow 0}1 \]

as long as x is not zero we can multiply both sides by x
\[ \lim_{x\rightarrow 0} f(x) = \lim_{x\rightarrow 0}x\]
as x goes to zero, write this as
\[ f(0)= 0 \]

Answer 3

Interesting. I knew that \[\lim_{x \rightarrow 0}c f(x) = c \lim_{x \rightarrow 0} f(x)\] (cool editor) but I hadn't suspected that c could be x. Learned another trick today....

Answer 4

After some time, that is a cool approach. Thanks for the lesson. That's why I looked for this place.

Answer 5

phi, I just realized that *that* wasn't what you did, but what you did was interesting too.

Answer 6

The key step is: lim(x>0) x * lim(x>0) f(x)/x = lim(x>0) x * lim(x>0) 1,

and so: lim(x>0) x * f(x)/x = lim(x>0) x * 1,

et cetera.