Prove lim f(x) as x approaches a is 0?

[jessjolt2]

Homework Statement

It is given that the limit as x->a of ( f(x)/(x-a) ) is 3. Prove using the espilon-delta theorem of limit that the limit as x->a of f(x) is 0.

The Attempt at a Solution

so it is known that: |( f(x)/(x-a) ) - 3| < E₁ when |x-a| < d₁
therefore:

|( f(x)/(x-a) ) - 3| ≤ |f(x)/(x-a)|+3 < E₁

|f(x)| < (E₁-3)|x-a|

(E₁-3)|x-a| ≤ C(E-3)

|x-a| ≤ 1=C

|f(x)| < (E₁-3) when d₁=min{ 1, (E-3) }


and we have to prove that |f(x)-0| < E when |x-a| < d

so can i just say that:

|f(x)-0| < E₂
|f(x)| < E₂ when |x-a| < d₂

So E₂=E₁-3 and d₂=d₁=min{ 1, (E-3) }

So |f(x)-0| < E₁-3 when |x-a| < d₁

is this right? i am not sure if i am making any logical sense in equating the two epsilons?


my intuition tells me that f(x) is zero at "a" because for f(x)/x-a to have a limit at "a", f(x) must be a factor of (x-a) so as to cancel out with the denominator. so if f(x)=(x-a)(p(x)), then f(a)=0, but i am not sure how to prove that using epsilon-delta, and i am not even sure if f(x) HAS to be a factor of (x-a)...sinx/x after all has a limit at 0 but sinx doesn't cancel with x.


help? :D

 

[HallsofIvy]

The denominator of f(x)/(x- 3) goes to 0 as x goes to 3. If f(x) goes to anything other than 0, then the limit of f(x)/(x-3), as x goes to 3, could not be finite.

It does NOT follow that f(3)= 0 because you are not told that f is continuous there. But that is not asked.

 

[jessjolt2]

The denominator of f(x)/(x- 3) goes to 0 as x goes to 3. If f(x) goes to anything other than 0, then the limit of f(x)/(x-3), as x goes to 3, could not be finite.

It does NOT follow that f(3)= 0 because you are not told that f is continuous there. But that is not asked.

why would it not be finite if f(x) did not go to 0?

specifically i want to know if my proof is right? my professor wants it to be proved using epsilon-delta, not just words..