Continuity of the elementary functions and standard limits

Summary

This lesson proves that the remaining elementary functions, the exponential, the logarithm, powers with real exponents, the inverse trigonometric functions and the hyperbolic functions, are continuous, with a single argument: a monotonic function whose values fill an interval cannot jump. With that continuity and the fundamental inequality of the exponential, the standard limits of the exponential and of the logarithm are proved, and the number $e$ is obtained as the limit of $(1 + x)^{\frac{1}{x}}$ . The lesson also resolves the indeterminate form $1^\infty$ and compares the exponential, powers and the logarithm at infinity.

Learning objectives

Prove that a monotonic function whose image is an interval is continuous, and deduce from this the continuity of the inverse and of the elementary functions.
Prove the standard limits of the exponential, of the logarithm and of powers.
Obtain the number $e$ as a limit and resolve the indeterminate form $1^\infty$ .
Compare the exponential, powers and the logarithm at infinity and near zero.

Monotonic functions without jumps

The lesson on limit and continuity proved that polynomials, roots, the absolute value and the trigonometric functions are continuous, and left the exponential, the logarithm and the inverse trigonometric functions pending. For them there is no bound as simple as $\abs{\sin x - \sin a} \leq \abs{x - a}$ . There is, however, a common argument: all of them are monotonic, and their values fill an interval without leaving gaps. A monotonic function can be discontinuous only through a jump, and a jump leaves a gap between the values.

TheoremMonotonic functions whose image is an interval

Let $f$ be a function defined on an interval $I$ and monotonic on it, and suppose that its image $f(I) = \left\{ f(x) : x \in I \right\}$ is an interval. Then $f$ is continuous on $I$ . By the lesson on one-sided limits, on each side of a point the monotonic function has a limit; if that limit were not the value, a gap would remain in the image. The figure shows an increasing function with a jump and the gap it leaves.

Proof

$\lim_{x \to a^-} f(x) = \dato{l}{L \leq f(a)}, \qquad f(c) \leq L$
It suffices to treat the increasing case: if $f$ is decreasing, $-f$ is increasing and its image is also an interval. Let $a$ be a point of $I$ other than its left endpoint, and take $c < a$ in $I$ . On $\ioo{c}{a}$ , the function is increasing and bounded by $f(a)$ ; by the theorem on the limit of a monotonic function, the left-hand limit exists and is the supremum of those values.
$x < a \rightarrow f(x) \leq L < y, \qquad x \geq a \rightarrow f(x) \geq f(a) > y$
Suppose that $L < f(a)$ . The midpoint $y$ between $L$ and $f(a)$ lies between the values $f(c)$ and $f(a)$ , so it belongs to the image, which is an interval. But no $x$ in $I$ attains it: if $x < a$ , the value does not exceed $L$ , and if $x \geq a$ , it does not fall below $f(a)$ . This is a contradiction.
$\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = \resaltar{f(a)}$
Therefore, $L = f(a)$ . In the same way, with the infimum and the corollary on the left endpoint from the same lesson, the right-hand limit equals $f(a)$ at every $a$ in $I$ other than its right endpoint. The two one-sided limits coincide with the value, and $f$ is continuous at $a$ ; at an endpoint of $I$ , the only side that exists suffices.

The hypothesis on the image is necessary: the integer part is monotonic and discontinuous at every integer, and its image, the set of integers, is not an interval. The continuity of inverse functions follows immediately from the theorem.

CorollaryContinuity of the inverse

Let $f$ be a function defined on an interval $I$ and strictly monotonic on it, and suppose that its image $J = f(I)$ is an interval. Then the inverse $f^{-1} \colon J \to I$ is continuous on $J$ . The inverse is monotonic in the same direction, and its image is the interval $I$ .

Proof

$f^{-1}(y) \geq f^{-1}(y') \rightarrow y \geq y', \qquad f^{-1}(J) = \resaltar{I}$
If $f$ is increasing and $y < y'$ in $J$ , the inverse cannot reverse the order: if $f^{-1}(y) \geq f^{-1}(y')$ held, applying $f$ would give $y \geq y'$ . Therefore $f^{-1}$ is increasing, and its image is $I$ ; the theorem gives continuity. The decreasing case is the same.