Exponential Tends to Zero and Infinity
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Contents
Theorem
Let $x \in \R$ be a real number.
Let $\exp x$ be the exponential of $x$.
Then:
- $\exp x \to +\infty$ as $x \to +\infty$
- $\exp x \to 0$ as $x \to -\infty$
Thus the exponential function has domain $\R$ and image $\left({0 \,.\,.\, +\infty}\right)$.
Proof
The exponential function approaches positive infinity as x approaches positive infinity
Let $M$ be a strictly positive real number.
Let $N$ be $\ln M$.
$N$ is real because $M > 0$.
From Exponential is Strictly Increasing:
- $\forall x: x > N \implies \exp x > \exp N = M$
Therefore:
- $\forall M \in \R_{>0} : \exists N \in \R : \forall x > N : \exp x > M$
The result follows from the definition of infinite limit at infinity.
$\blacksquare$
The exponential function approaches 0 as x approaches negative infinity
Let $\epsilon$ be a strictly positive real number.
Let $c$ be $\ln \epsilon$.
Let $x$ be any real number that smaller than $c$.
From Exponential of Real Number is Strictly Positive:
- $\exp x > 0 \implies \left\vert{\exp x}\right\vert = \exp x$
From Exponential is Strictly Increasing:
- $\left\vert{\exp x - 0}\right\vert = \exp x < \exp c = \epsilon$
Therefore:
- $\forall \epsilon \in \R_{>0} : \exists c \in \R : \forall x < c : \left\vert{\exp x - 0}\right\vert < \epsilon$
From the definition of limit involving infinity, the result follows.
$\blacksquare$
The exponential function has domain $\R$ and image $\left({0 \,.\,.\, +\infty}\right)$
the above uses this result
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We have that the Exponential is Strictly Increasing.
From above, $\displaystyle \lim_{x \to \infty} \exp x = \infty$
From above, $\displaystyle \lim_{x \to -\infty} \exp x = 0$
Hence the result.
$\blacksquare$
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 14.4$
