Let
𝑓
,
𝑔
be positive, continuous functions with domain
𝑅
.

Let
𝐿
=
lim
𝑥
→
∞
𝑓
(
𝑥
)
𝑔
(
𝑥
)
.

Assume
∫
1
∞
𝑓
(
𝑥
)
𝑑
𝑥
is convergent.

Which of the following statements must be true?

Select all the correct answers.

Question

Let 
𝑓
,
𝑔
  be positive, continuous functions with domain 
𝑅
  .

Let 
𝐿
=
lim
𝑥
→
∞
𝑓
(
𝑥
)
𝑔
(
𝑥
)
 .

Assume 
∫
1
∞
𝑓
(
𝑥
)
𝑑
𝑥
 is convergent.

Which of the following statements must be true?

Select all the correct answers.

BlackTom AI · Accepted Answer

We start by parsing the given setup: f and g are positive, continuous on R, and L = lim_{x→∞} f(x)/g(x). It is also given that ∫_1^∞ f(x) dx converges. We will evaluate each statement in light of these facts and general limit comparison ideas.

Option 1: IF L DNE, THEN ∫_1^∞ g(x) dx is not convergent.
This claim tries to link the nonexistence of the limit of f/g to the divergence of the integral of g. However, the absence of a limit does not force divergence of ∫ g. It is entirely possible for f and g to behave in ways that prevent a limit from existing while still having ∫ g converge or diverge independently. For example, one could have oscillatory behavior in f/g that prevents a limit, yet g's tail could be integrable. Therefore, this statement is not guaranteed to be true in all scenarios; it is not a must.

Option 2: IF L exists, THEN ∫_1^∞ g(x) dx is convergent.
Here we consider the case when the limit L exists and is finite. If L > 0, then f ~ L g for large x, so convergence of ∫ f implies convergence of ∫ g (since multiplying by a positive constant preserves convergence of improper integrals). If L = 0, however, f is little-o of g, meaning f is much smaller than g asymptotically. In that case, the convergence of ∫ f does not force the convergence of ∫ g; g could be large enough on a tail to cause divergence even though f remains small enough to be integrable. Because the statement would have to hold even when L = 0, it is not guaranteed to be true. Thus this option is not must.

Option 3: IF L = 3, THEN ∫_1^∞ g(x) dx is convergent.
This is a straightforward application of the limit comparison principle. If L = 3 (a positive finite limit), then for large x we have f(x) ≈ 3 g(x). Since ∫ f(x) dx converges, the comparison f ≈ 3 g implies ∫ g must also converge; multiplying a function by a positive constant does not affect convergence. Therefore, this statement must be true.

Option 4: IF L is ∞, THEN ∫_1^∞ g(x) dx is convergent.
If L = ∞, then f(x)/g(x) → ∞, so for any large M there exists X such that for x > X we have f(x) ≥ M g(x). In particular, for large x, f(x) dominates g(x) by an unbounded factor. Since ∫_1^∞ f(x) dx converges, the tail integral ∫_X^∞ f(x) dx is finite. But for x > X we have f(x) ≥ M g(x) with arbitrarily large M, forcing ∫_X^∞ g(x) dx to be bounded above by (1/M) ∫_X^∞ f(x) dx, which tends to zero as M grows. This implies ∫_1^∞ g(x) dx must converge as well. Hence this statement is also must be true.

Let 𝑓 , 𝑔 be positive, continuous functions with domain 𝑅 . Let 𝐿 = lim 𝑥 → ∞ 𝑓 ( 𝑥 ) 𝑔 ( 𝑥 ) . Assume ∫ 1 ∞ 𝑓 ( 𝑥 ) 𝑑 𝑥 is convergent. Which of the following statements must be true? Select all the correct answers.Multiple choice

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