Sequential Characterization of Limit at Minus Infinity of Real Function

Theorem

Let $f : \R \to \R$ be a real function.

Let $L$ be a real number.

Then:

$\ds \lim_{x \to -\infty} \map f x = L$

if and only if:

for all real sequences $\sequence {x_n}_{n \mathop \in \N}$ with $x_n \to -\infty$ we have $\map f {x_n} \to L$

where:

$\ds \lim_{x \mathop \to -\infty} \map f x$ denotes the limit at $-\infty$ of $f$.

Corollary

$\ds \lim_{x \to -\infty} \map f x = L$

if and only if:

for all decreasing real sequences $\sequence {x_n}_{n \mathop \in \N}$ with $x_n \to -\infty$ we have $\map f {x_n} \to L$

where:

$\ds \lim_{x \mathop \to -\infty} \map f x$ denotes the limit at $-\infty$ of $f$.

Proof

Necessary Condition

Suppose that:

$\ds \lim_{x \to -\infty} \map f x = L$

Let $\sequence {x_n}_{n \mathop \in \N}$ be a real sequence with $x_n \to -\infty$.

Let $\epsilon > 0$.

From the definition of limit at $-\infty$, we have:

there exists $M > 0$ such that for all $x < -M$ we have $\size {\map f x - L} < \epsilon$.

Since $\sequence {x_n}_{n \mathop \in \N}$ diverges to $-\infty$, we have:

there exists $N \in \N$ such that $x_n < -M$ for all $n \ge N$.

That is, for all $n \ge N$, we have:

$\size {\map f {x_n} - L} < \epsilon$

Since $\epsilon$ was arbitrary, we have:

$\map f {x_n} \to L$

Since $\sequence {x_n}_{n \mathop \in \N}$ was arbitrary:

for all real sequences $\sequence {x_n}_{n \mathop \in \N}$ with $x_n \to -\infty$ we have $\map f {x_n} \to L$.

So, if:

$\ds \lim_{x \to -\infty} \map f x = L$

then:

for all real sequences $\sequence {x_n}_{n \mathop \in \N}$ with $x_n \to -\infty$ we have $\map f {x_n} \to L$.

$\Box$

Sufficient Condition

We prove the contrapositive.

Suppose that it is not the case that:

$\ds \lim_{n \mathop \to \infty} \map f x = L$

We show that:

there exists a real sequence $\sequence {x_n}_{n \mathop \in \N}$ with $x_n \to -\infty$ but such that $\sequence {\map f {x_n} }_{n \mathop \in \N}$ does not converge to $L$.

Then:

there exists some $\epsilon > 0$ such that for all $M > 0$ there exists $x < -M$ such that $\size {\map f x - L} \ge \epsilon$.

We construct $\sequence {x_n}_{n \mathop \in \N}$ inductively.

Pick $x_1$ such that $x_1 < -1$ and:

$\size {\map f {x_1} - L} \ge \epsilon$

Given $x_1, x_2, \ldots, x_{n - 1} < 0$, pick $x_n$ so that:

$x_n < \min \set {-n, x_1, x_2, \ldots, x_{n - 1} } \le x_{n - 1}$

and:

$\size {\map f {x_n} - L} \ge \epsilon$

Then $\sequence {x_n}_{n \mathop \in \N}$ is decreasing.

We also have:

$x_n < -n$ for each $n$

so $\sequence {x_n}_{n \mathop \in \N}$ is unbounded below.

From Unbounded Monotone Sequence Diverges to Infinity: Decreasing, we therefore have:

$x_n \to -\infty$

Since:

$\size {\map f {x_n} - L} \ge \epsilon$

there exists no $N \in \N$ such that:

$\size {\map f {x_n} - L} < \epsilon$

for each $n \ge N$.

So:

$\sequence {\map f {x_n} }_{n \mathop \in \N}$ does not converge to $L$.

We conclude that if it is not the case that:

$\ds \lim_{x \mathop \to -\infty} \map f x = L$

then:

there exists a real sequence $\sequence {x_n}_{n \mathop \in \N}$ with $x_n \to -\infty$ but such that $\sequence {\map f {x_n} }_{n \mathop \in \N}$ does not converge to $L$.

So, if:

$\ds \lim_{x \mathop \to -\infty} \map f x = L$

then:

for all real sequences $\sequence {x_n}_{n \mathop \in \N}$ with $x_n \to -\infty$ we have $\map f {x_n} \to L$.

$\blacksquare$