https://r2.erweima.ai/mathbot/image/17243908235155grb8fhb.pdf

esto es la teoria, quiero que vuelvas a resolver los ejercicios anteriores, pero basate en esta teoria, además cuando resuelveas no uses palabras, solo resuelvelo y dame el procedimiento sin palabras

1a) \( f(x) = x^2 - 4x + 3 \)

\[
f(2) = 2^2 - 4(2) + 3 = 4 - 8 + 3 = -1
\]

\[
\lim_{{x \to 2^-}} f(x) = \lim_{{x \to 2^+}} f(x) = -1
\]

\( \text{Continua en } x = 2 \)

1b) \( f(x) = \frac{x^2 - 4}{x - 2} \)

\[
\text{Simplificación: } f(x) = \frac{(x-2)(x+2)}{x-2} = x + 2, \text{para } x \neq 2
\]

\[
\lim_{{x \to 2}} f(x) = 4
\]

\( \text{Discontinua en } x = 2 \)

1c) \( g(x) = \frac{x^2 - 3}{x^2 - 4} \)

\[
g(2) = \frac{4 - 3}{4 - 4} = \frac{1}{0} \text{(Indefinido)}
\]

\( \text{Discontinua en } x = 2 \)

2a) \( f(x) = \left\{ \begin{array}{ll} 
9 - x^2 & \text{si } x \leq 2 \\ 
x + 3 & \text{si } x > 2 
\end{array} \right. \)

\[
f(2) = 9 - 4 = 5
\]

\[
\lim_{{x \to 2^-}} f(x) = 5, \quad \lim_{{x \to 2^+}} f(x) = 5
\]

\( \text{Continua en } x = 2 \)

3) \( f(x) = \left\{ \begin{array}{ll} 
2x + 3 & \text{si } x < 4 \\ 
kx + 5 & \text{si } x \geq 4 
\end{array} \right. \)

\[
\lim_{{x \to 4^-}} f(x) = 2(4) + 3 = 11
\]

\[
\lim_{{x \to 4^+}} f(x) = 4k + 5 = 11 \implies k = 1.5
\]

Explore how limits and continuity are applied in various functions through detailed examples. This includes evaluating limits at specific points and understanding continuity criteria. Suitable for advanced high school students.

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