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## 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| < E

_{1}when |x-a| < d

_{1}

therefore:

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

_{1}

|f(x)| < (E

_{1}-3)|x-a|

(E

_{1}-3)|x-a| ≤ C(E-3)

|x-a| ≤ 1=C

|f(x)| < (E

_{1}-3) when d

_{1}=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| < E

_{2}

|f(x)| < E

_{2}when |x-a| < d

_{2}

So E

_{2}=E

_{1}-3 and d

_{2}=d

_{1}=min{ 1, (E-3) }

So |f(x)-0| < E

_{1}-3 when |x-a| < d

_{1}

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 doesnt cancel with x.

help? :D