Question

prove that lim f(cx) = lim f(x) as x approaches to 0

Answer 1

do you have the definition of a limit ?

Answer 2

hey the question is right, it is a exam question.

Answer 3

ok, i think they want us to use the definition of a limit

Answer 4

precise definition of limit

Answer 5

https://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/preciselimdirectory/PreciseLimit.html

Answer 6

also you have to assume that a limit exists as x - > 0

Answer 7

there might be an easier way,

Answer 8

use composition of limit theorem

Answer 9

since
as x ->0 , c*x -> 0 for any constant c
lim f( c*x) should equal to lim f (x)
x -> 0 x -> 0

Answer 10

it makes sense intuitively.

Answer 11

if we assume that
lim f(cx) = 0 exists as x->0,
then that means formally
for any positive epsilon
if | x - 0 | < delta then | f(cx) - L | < epsilon

Answer 12

but | x- 0 | < delta is equivalent to c * | x - 0 | < c*delta

Answer 13

since we know that the limit f(cx) exists as x -> 0 . that means we know that

for any e
| x - 0 | < d
| f(c*x) - L | < e .

now from this what can you show

Answer 14

another approach is to assume the two limits are not equal, and derive a contradiction

Answer 15

ok i have a new idea

Answer 16

also notice that these steps are bi-directional, so if you start with the proof by assuming lim f(u) as u->0 exists, then you can get that it equals to lim f(cx) as x ->0

Answer 17

i shouldnt have erased that

Answer 18

@perl

Answer 19

We know that lim f(cx) exists , as x->0. Lets call this limit L
This implies,
" for any e >0 there exists d >0 such that
if | x - 0 | < d , then | f( cx ) - L | < e "
but | x - 0 | < d
is equivalent to c*| x - 0 | < c*d .
Now call c*d a new delta, d '.
Substituting above we have
" for any e > 0 there exists d ' > 0 such that
if | c*x - 0 | < d ' , then | f( cx ) - L | < e . "
This implies limit f(u) = L , as u approaches zero, where u = c*x.