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can you answer these 2 questions using purely epsilon delta definition of limit for everything

Let's tackle these two questions using the epsilon-delta definition of limits. I'll first give a brief overview of how the epsilon-delta definition works, and then we'll address the true/false statements.

### Epsilon-Delta Definition of Limit:
For a function \( f(x) \), we say that:
\[
\lim_{x \to a} f(x) = L
\]
if for every \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that whenever \( 0 < |x - a| < \delta \), we have \( |f(x) - L| < \epsilon \).

### Question (a):
**Statement**: IF \( \lim_{x \to a} [f(x) \cdot g(x)] \) exists and \( \lim_{x \to a} g(x) \) exists, THEN \( \lim_{x \to a} f(x) \) exists.

Let's break it down:
- Let \( L_1 = \lim_{x \to a} [f(x) \cdot g(x)] \) and \( L_2 = \lim_{x \to a} g(x) \).
- We need to show whether \( \lim_{x \to a} f(x) \) exists based on the existence of these two limits.

### Proof/Disproof:
To determine if this is true, we need to investigate whether knowing \( \lim_{x \to a} [f(x) \cdot g(x)] \) and \( \lim_{x \to a} g(x) \) guarantees the existence of \( \lim_{x \to a} f(x) \).

A counterexample disproves this:
- Let \( f(x) = \frac{1}{x - a} \) and \( g(x) = x - a \). Notice:
  \[
  f(x) \cdot g(x) = 1 \quad \text{(constant function)}.
  \]
  Hence, \( \lim_{x \to a} [f(x) \cdot g(x)] = 1 \).
- Meanwhile, \( \lim_{x \to a} g(x) = 0 \).
- However, \( \lim_{x \to a} f(x) = \lim_{x \to a} \frac{1}{x - a} \) **does not exist** because the function blows up to infinity as \( x \to a \).

Thus, the statement is **false**. The existence of the product limit \( \lim_{x \to a} [f(x) \cdot g(x)] \) and \( \lim_{x \to a} g(x) \) does not guarantee that \( \lim_{x \to a} f(x) \) exists.

### Question (b):
**Statement**: IF \( \lim_{x \to a} f(x) \) does not exist and \( \lim_{x \to a} g(x) \) does not exist, THEN \( \lim_{x \to a} [f(x) + g(x)] \) does not exist.

### Proof/Disproof:
We need to analyze whether the non-existence of both individual limits implies that the sum of the functions does not have a limit.

This is **false**, as we can construct a counterexample:
- Let \( f(x) = -\frac{1}{x - a} \) and \( g(x) = \frac{1}{x - a} \).
- Both \( \lim_{x \to a} f(x) \) and \( \lim_{x \to a} g(x) \) do not exist because they go to \( -\infty \) and \( +\infty \), respectively.
- However, \( f(x) + g(x) = 0 \), which is constant.
- Hence, \( \lim_{x \to a} [f(x) + g(x)] = 0 \), which exists.

Thus, the statement is **false**. Even if both \( \lim_{x \to a} f(x) \) and \( \lim_{x \to a} g(x) \) do not exist, their sum can still have a finite limit.

---

Let me know if you'd like more detailed steps on these proofs or further clarifications!

### Follow-up Questions:
1. How does the epsilon-delta definition relate to product or sum limits in general?
2. What are other possible counterexamples where limits behave unexpectedly?
3. How can we use the epsilon-delta definition to prove the existence of a limit rigorously?
4. How can limits that approach infinity or fail to exist affect the behavior of related functions?
5. Are there cases where the sum of non-existent limits might still fail to exist, despite these counterexamples?

### Tip:
When tackling limit proofs, always look for counterexamples before assuming a statement is true. These are often the simplest way to disprove general claims.

Let a ∈ R. Let f and g be two functions defined on R. Is each of the following claims true or false? Prove your answer using the epsilon-delta definition of a limit. (a) If lim_{x→a} [f(x) · g(x)] exists and lim_{x→a} g(x) exists, then lim_{x→a} f(x) exists. (b) If lim_{x→a} f(x) does not exist and lim_{x→a} g(x) does not exist, then lim_{x→a} [f(x) + g(x)] does not exist.

Explore whether the existence of product or sum limits guarantees the existence of individual limits using the epsilon-delta definition. Includes counterexamples for two cases: (a) the existence of a product limit and (b) the non-existence of individual limits. Suitable for undergraduate mathematics students.

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