Limits and Continuity

It implies that $\lim_{{t \to a^-}} h(t) = \lim_{{t \to a^+}} h(t) = h(a)$.

It infers only that there are no holes or breaks in its graph near $t=a$.

It requires solely that $h(a)$ exists and does not concern itself with limits around $t=a$.

It suggests that only one-sided limits must exist and be finite around $t=a$.

4

$\frac{4}{0}$

Undefined

0

f(x) = \frac{1}{x}

f(x) = \frac{\sqrt{x}}{x}

f(x) = x^2 - x

f(x) = \frac{x-1}{x^2-x}

Since the denominator becomes zero at 8, continuity breaks there.

No condition needed since $k(z)$ simplifies to a constant, therefore uninterrupted wherever it's defined

Must exclude 8 and its vicinity to prevent interruption caused by division by zero.

Z cannot approach or reach 8.

Discontinuities are present in immediate vicinity of $G(A)$.

G(a) might have an asymptotic behavior.

G(a) must be finite.

There is possible oscillation near $G(A)$.

Three points because each interval may contribute one potential point of discontinuity.

One point if $-5$ poses an issue but not necessarily others.

Infinitely many within intervals if expressions aren't polynomial or rational.

Two points since only boundaries between intervals require checking for continuity.

f(x) = |x|

f(x) = \sqrt{x}

f(x) = \frac{1}{x}

f(x) = x^2

Check for any polynomials

Draw a graph and find the open points

Determine if there are any holes and any asymptotes /jumps

Factor out any like variables

Trigonometric functions

Polynomial functions

Rational functions

Exponential functions

f(x) = \frac{1}{x-6}

f(x) = \frac{x-4}{x^2-4x}

f(x) = \frac{2x-10}{x^2+x-30}

f(x) = \frac{2x-24}{2x+14}