Need help with the following proof:
prove that if lim x->c 1/f(x)= 0 then
lim x->c f(x) does not exist.I think i need to use the delta epsilon definition i am not sure how to set it up.
In general,
A implies B
is equivalent to
Not(B) implies not (A)
This means that the statement:
lim x->c 1/f(x)= 0 ------->
lim x->c f(x) does not exist,
is equivalent to the statement:
lim x->c f(x) exists ------->
lim x->c 1/f(x) is not zero
Now, we can make the statement more precise.
If lim x->c f(x) exists, then there is some rela number y such that:
lim x->c f(x) = y
Then, it is easy to prove that if y is not zero, we have:
lim x->c 1/f(x) = 1/y
If y = 0, then:
lim x->c 1/f(x) does not exist.
So, we always have that the limit of
1/f(x) is not equal to zero.
To prove that if the limit of 1/f(x) as x approaches c is 0, then the limit of f(x) as x approaches c does not exist, you can indeed use the epsilon-delta definition of limits. Here's how you can set it up:
We want to show that if lim x→c 1/f(x) = 0, then lim x→c f(x) does not exist.
By the definition of a limit, for 1/f(x) to tend to 0 as x approaches c, we can say that for any ε > 0, there exists a δ > 0 such that if 0 < |x - c| < δ, then |1/f(x) - 0| < ε.
Now, let's assume that lim x→c f(x) exists. This means that for some L, for any ε' > 0, there exists a δ' > 0 such that if 0 < |x - c| < δ', then |f(x) - L| < ε'.
Let's choose ε' = ε/2. Since lim x→c f(x) exists, we know that there exists a δ' > 0 that satisfies the properties mentioned above.
Now, let's choose a specific x such that 0 < |x - c| < δ, where δ is the one that corresponds to ε from the definition of the limit of 1/f(x) as x → c.
For this x, we have:
|1/f(x) - 0| < ε (from the definition of the limit of 1/f(x) as x → c)
|1/f(x)| < ε
1/|f(x)| < ε
|f(x)| > 1/ε (since f(x) ≠ 0)
Now, let's see what happens when we choose this x to satisfy the properties for the limit of f(x) as x → c:
|f(x) - L| < ε' = ε/2
Combining the inequalities, we have:
|f(x)| - |L| < |f(x) - L| < ε/2
|f(x)| - |L| < ε/2
|f(x)| < ε/2 + |L|
Comparing this inequality to |f(x)| > 1/ε, we have:
1/ε < ε/2 + |L|
Rearranging the inequality, we get:
1/ε - ε/2 < |L|
Now, we have a contradiction because the magnitude of L must be greater than or equal to 0. However, for any ε > 0, we can always find a positive ε for which this inequality does not hold. Thus, lim x→c f(x) cannot exist if lim x→c 1/f(x) = 0.
Therefore, we have proven that if the limit of 1/f(x) as x approaches c is 0, then the limit of f(x) as x approaches c does not exist.