Sequential Characterization of Limit at Positive Infinity of Real Function

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

Let $L$ be a real number.

Then:


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




 * 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:


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

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 infinity, 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 infinity, 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 $\map f {x_n} \to L$.

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


 * $x_n > \max \set {n, x_1, x_2, \ldots, x_{n - 1} } \ge x_{n - 1}$

and:


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

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

We also have:


 * $x_n > n$ for each $n$

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

From Unbounded Monotone Sequence Diverges to Infinity, 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:


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

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:


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

then:


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