Limits, continuity, differentiability (HL)

Continuity is a necessary but insufficient condition for differentiability. Both are based on limits

This continues the discussions from completeness of real numbers, intermediate value theorem (HL), and limits from table of values, which will not be repeated here.

Contents

Limits at infinity

Limits at infinity include both x→∞x \to \infty and x→−∞x \to -\infty. The two limits lead to a maximum of two horizontal asymptotes for a function. Limits of sequences are only very briefly touched upon in this course, but they follow the same ideas.

  1. Oscillation that maintains its amplitude is diverging (limit does not exist). For example
lim⁡x→∞sin⁡(x)    dne\lim_{x\to \infty} \sin(x) \;\; \text{dne}
  1. If a limit at infinity exists, then there is a horizontal asymptote. The function may approach the asymptote from above, approach from below, or oscillate towards it at decreasing amplitudes.

If the limit at (positive) infinity exists, the function (or sequence) converges, else it diverges.

Limit rules

Some common and useful properties of limits.

lim⁡x→af(x)g(x)f(x)=lim⁡x→ag(x)\lim_{x\to a} \frac{f(x)g(x)}{f(x)} = \lim_{x\to a} g(x)
lim⁡x→akg(x)=klim⁡x→ag(x)\lim_{x\to a} kg(x) = k\lim_{x\to a}g(x)

For continuous function ff, and finite g(x)g(x)

lim⁡x→af(x)=f(a)\lim_{x\to a} f(x) = f(a)
lim⁡x→af(g(x))=f(lim⁡x→ag(x))\lim_{x\to a} f(g(x)) = f\left(\lim_{x\to a} g(x)\right)

Operations are not defined or available for undefined. This means you can splits a single limit only into a sum or product, LL of two finite limits. You cannot break into limits involving undefined or indeterminate forms.

lim⁡x→af(x)+lim⁡x→ag(x)=L  ⟹  lim⁡x→a(f(x)+g(x))=L\lim_{x\to a} f(x) + \lim_{x\to a} g(x) = L \implies \lim_{x\to a} (f(x) + g(x)) = L
lim⁡x→af(x)⋅lim⁡x→ag(x)=L  ⟹  lim⁡x→a(f(x)⋅g(x))=L\lim_{x\to a} f(x) \cdot \lim_{x\to a} g(x) = L \implies \lim_{x\to a} (f(x) \cdot g(x)) = L

Example: Evaluate

lim⁡x→∞x+lim⁡x→∞(−x)\lim_{x\to \infty} x + \lim_{x\to \infty} (-x)

Undefined ■\qed

Because both individual limits are undefined.

Yeah limits are weird.

Example: Evaluate

lim⁡x→0+1x⋅lim⁡x→0+x\lim_{x\to 0^+} \frac 1x \cdot \lim_{x\to 0^+} x

Undefined ■\qed

Because the first limit is undefined.

Example: Evaluate

lim⁡x→πtan⁡xsin⁡x\lim_{x\to \pi} \frac{\tan x}{\sin x}
lim⁡x→πtan⁡xsin⁡x=lim⁡x→πsin⁡xsin⁡xcos⁡x=lim⁡x→π1cos⁡x=−1■\begin{align*} \lim_{x\to \pi} \frac{\tan x}{\sin x} &= \lim_{x\to \pi} \frac{\sin x}{\sin x \cos x} \\ &= \lim_{x\to \pi} \frac{1}{\cos x} \\ &= -1 \qed \end{align*}

Normal algebraic rules apply within a limit.

Continuity

A function f(x)f(x) is continuous at x=x0x = x_0 if and only if

lim⁡x→x0f(x)=f(x0)\lim_{x\to x_0} f(x) = f(x_0)

Since real numbers are continuous, it is more interesting to see types of discontinuity.

vertical asymptotes

Vertical asymptotes are also known as infinite discontinuities, arising from the fact that ±∞\pm \infty are not real numbers.

jump discontinuity

Here, the two one-sided limits converge to different values, meaning that lim⁡x→x0f(x)\displaystyle \lim_{x\to x_0} f(x) is not defined, even though f(x)f(x) may be defined.

removable discontinuity

In this case, f(x0)f(x_0) is defined at a different point than lim⁡x→x0f(x)\displaystyle \lim_{x\to x_0} f(x). The discontinuity can be taken out by simply redefining f(x0)f(x_0) to match the limit.

Differentiability

A continuous function is differentiable at x=x0x = x_0 if and only if

lim⁡h→0f(x+h)−f(x)h=f′(x)\lim_{h\to 0} \frac{f(x+h) - f(x)}{h} = f^\prime(x)

is defined at x=x0x = x_0.

Not differentiable means either the two one-side limits do not approach the same number (eg x=0x = 0 on x↦∣x∣x \mapsto \lvert x\rvert) or a vertical tangent (eg x=0x = 0 on x↦x13x \mapsto x^{\frac13} using the real-valued root).

Example: Find

 d dxx3\frac{\d }{\d x} x^3

from first principles.

L=lim⁡h→0f(x+h)−f(x)h=lim⁡h→0(x+h)3−x3h=lim⁡h→0(x3+3hx2+3h2x+h3)−x3h=lim⁡h→03hx2+3h2x+h3h=lim⁡h→0(3x2+3hx+h2)\begin{align*} L &= \lim_{h\to 0} \frac{f(x+h) - f(x)}{h} \\ &= \lim_{h\to 0} \frac{(x+h)^3 - x^3}{h} \\ &= \lim_{h\to 0} \frac{(x^3+3hx^2 + 3h^2x + h^3) - x^3}{h} \\ &= \lim_{h\to 0} \frac{3hx^2 + 3h^2x + h^3}{h} \\ &= \lim_{h\to 0} {(3x^2 + 3hx + h^2)} \end{align*}

Since lim⁡h→03x2\displaystyle {\lim_{h\to 0}3x^2}, lim⁡h→03hx\displaystyle {\lim_{h\to 0}3hx}, lim⁡h→0h2\displaystyle {\lim_{h\to 0}h^2} are all finite, the single limit can be split into three limits.

L=lim⁡h→03x2+lim⁡h→03hx+lim⁡h→0h2=3x2+0+0=3x2■\begin{align*} L &= \lim_{h\to 0}3x^2 + \lim_{h\to 0}3hx + \lim_{h\to 0}h^2 \\ &= 3x^2 + 0 + 0 \\ &= 3x^2 \qed\end{align*}

The fact that 3x23x^2 is continuous was also implicitly used.

Final remarks

You do not have to prove a function is continuous or differentiable. You only have to identify discontinuities, make piecewise functions continuous (and possibly differentiable), and differentiate polynomials from first principles.