# Confusion regarding a proof for an infinite limit property.

1. Mar 13, 2013

### InaudibleTree

Assume for some real number L and c

$\displaystyle\lim_{x\rightarrow c} f(x) = ∞$ and $\displaystyle\lim_{x\rightarrow c} g(x) = L$

We must prove

$\displaystyle\lim_{x\rightarrow c} [f(x) + g(x)] = ∞$

Let $M > 0$. We know $\displaystyle\lim_{x\rightarrow c} f(x) = ∞$. Thus,

there exists $δ_1>0$ such that if $0 < |x - c| < δ_1$ we have,

$f(x) > M - L + 1$.

Also, we know $\displaystyle\lim_{x\rightarrow c} g(x) = L$. Thus,

there exists $δ_2 > 0$ such that if $0 < |x - c| < δ_2$ we have,

$0 < |g(x) - L| < 1 →→ -1 < g(x) - L < 1 →→ L - 1 < g(x) < L + 1$

Let $δ = min(δ_1,δ_2)$. And so if $0 < |x - c| < δ$ we will have both,

$f(x) > M - L + 1$ and $g(x) > L - 1$

Thus,

$f(x) + g(x) > M - L + 1 + L - 1 = M$

Now what confuses me is how the proof can get to the point

$f(x) > M - L + 1$

without the assumption that $L > 0$.

Afterall following from the definition of an infinite limit

$M - L + 1 > 0$ and $M > 0$. Right?

Im sure I must be confusing something here. Any help would be appreciated.

2. Mar 13, 2013

### Stephen Tashi

The proof doesn't deduce that $f(x) > M - L + 1$ from the step above your question.

The fact that $f(x) > M - L + 1$ is established earlier in the proof. It is established from the fact that f(x) can be made larger than any given number, so in particular it can be made larger than the number M - L + 1 by the appropriate choice of a delta. The proof says to make an appropriate choice.

3. Mar 13, 2013

### InaudibleTree

Im sorry I should of structured my post better. I knew the inequality given was established earlier in the proof.

I reread the definition of infinite limits. I think what you are saying makes sense.

Im really just allowed to choose an $M>0$(in this case $M -L + 1$) and at some point(s) $f(x) > M$. Afterall $f$ is defined at every real number in some open interval containing c.

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