[dario_oasis]

Me dicen si hice bien este ejercicio?


\( \displaystyle\lim_{x \to{}1}{\displaystyle\frac{x+1}{2(x-2)}}=-1 \)

\( \forall{\in{>0}},\exists{\delta}>0,\delta(\in{})/0</x-x0/<\delta\Rightarrow{\left |{f(x)-l}\right |}<\in{} \)

\( \left |{\displaystyle\frac{X+1+2X-4}{2x-4}}\right | \)

\( \left |{\displaystyle\frac{3X-3}{2(X-2)}}\right | \)

\( \left |{\displaystyle\frac{3(X-1)}{2(X-2)}}\right | \)


proponemos \( \delta=1/4 \)

\( -1/4< x-1 <1/4 \)

\( 3/4<X<5/4 \)

\( 3/4-2<X-2<5/4-2 \)

\( -5/4<X-2<-3/4 \)


\( 3/4<\left |{X-2}\right |<5/4 \)

\( 4/3>\displaystyle\frac{1}{\left |{X-2}\right |}>4/5 \)


\( 3/2.\delta.4/3<\in{} \)

\( 2\delta<\in{} \)

\( \delta=\displaystyle\frac{\in{}}{2} \)

Luego \( \delta=min(\displaystyle\frac{\in{}}{2},1/4) \)

[Fernando Revilla]

Luego \( \delta=min(\displaystyle\frac{\in{}}{2},1/4) \)

Lo tienes bien. Como prueba (la vuelta), y sin mirar lo que tú has hecho (la ida):

Caso I. \( \delta=\dfrac{1}{4}\le \dfrac{\epsilon}{2} \). Entonces,

          \( \boxed{\left|x-1\right|<\delta}\Rightarrow \left|x-1\right|<1/4\Rightarrow -1/4<x-1<1/4\Rightarrow -5/4<x-2<-3/4 \)
          \( \Rightarrow \left|x-2\right|>3/4\Rightarrow \dfrac{1}{\left|x-2\right|}<\dfrac{4}{3}\Rightarrow \dfrac{3\left|x-1\right|}{2\left|x-2\right|}<\underbrace{\dfrac{3(\epsilon/2)}{2}\cdot \dfrac{4}{3}}_{=\;\epsilon}\Rightarrow \boxed{\dfrac{3\left|x-1\right|}{2\left|x-2\right|}<\epsilon} \)

Caso II. \( \delta=\dfrac{\epsilon}{2}<\dfrac{1}{4} \). Entonces,

          \( \boxed{\left|x-1\right|<\delta}\Rightarrow \left|x-1\right|<\epsilon/2\;\wedge\; \dfrac{1}{\left|x-2\right|}<\dfrac{4}{3}\Rightarrow \dfrac{3\left|x-1\right|}{2\left|x-2\right|}<\underbrace{\dfrac{3(\epsilon/2)}{2}\cdot \dfrac{4}{3}}_{=\;\epsilon}\Rightarrow \boxed{\dfrac{3\left|x-1\right|}{2\left|x-2\right|}<\epsilon} \)

P.D. Si tecleas [tex]\delta=\min\left(\displaystyle\frac{\epsilon/2}{2},1/4\right)[/tex] aparecerá

          \( \delta=\min\left(\displaystyle\frac{\epsilon/2}{2},1/4\right) \),

más estético que tu \( \delta=min(\displaystyle\frac{\in{}}{2},1/4) \).