# User:"juan carlos arango"

## LIMITES

1)${\displaystyle \lim _{x\to -2}{\frac {x^{3}+2x^{2}+1-1}{x+2}}}$

=${\displaystyle \lim _{x\to -2}{\frac {x^{3}+2x^{2}}{x+2}}}$

=${\displaystyle \lim _{x\to -2}{\frac {x^{2}(x+2)}{x+2}}}$

=${\displaystyle \lim _{x\to -2}\ x^{2}}$

=${\displaystyle \ 4}$

BY JUAN CARLOS ARANGO

2)${\displaystyle \lim _{x\to 1}{\frac {x^{2}-1}{x^{3}-1}}}$

=${\displaystyle \lim _{x\to 1}{\frac {(x-1)(x+1)}{(x-1)(x^{2}+x+1}}}$

=${\displaystyle \lim _{x\to 1}{\frac {x+1}{x^{2}+x+1}}}$

=${\displaystyle {\frac {2}{1+1+1}}}$

=${\displaystyle {\frac {2}{3}}}$

BY JUAN CARLOS ARANGO

3)${\displaystyle \lim _{x\to 1}{\frac {t^{3}-t}{t^{3}-1}}}$

=${\displaystyle \lim _{x\to 1}{\frac {t(t^{-}1)}{(t-1)(t^{2}+t+1}}}$

=${\displaystyle \lim _{x\to 1}{\frac {t(t+1)(t-1)}{t-1(x^{2}+x+1)}}}$

=${\displaystyle {\frac {1(1+1)}{1^{2}+1+1}}}$

=${\displaystyle {\frac {2}{3}}}$

BY JUAN CARLOS ARANGO

4)${\displaystyle \ y={\frac {4x^{3}-2x}{7}}}$

${\displaystyle \ y'={\frac {12x^{2}-2(7)-0(4x^{3}-2x)}{(7^{2})}}}$
${\displaystyle \ y'={\frac {12x^{2}-2(7)}{(7^{2})}}}$
${\displaystyle \ y'={\frac {12x^{2}-2}{7}}}$

BY JUAN CARLOS ARANGO

5)${\displaystyle \ y={\frac {4-3x}{2+x}}}$

${\displaystyle \ y'={\frac {-3(2+x)-(4-3x)(1)}{(2+x)^{2}}}}$
${\displaystyle \ y'={\frac {-6-3x-4+3x}{(2+x)^{2}}}}$
${\displaystyle \ y'={\frac {-10}{(2+x)^{2}}}}$

BY JUAN CARLOS ARANGO

6) ${\displaystyle \ y={x^{3}-x^{2}+2x}}$

${\displaystyle \ y'={3x^{2}-2x+2}}$

BY JUAN CARLOS ARANGO

7)${\displaystyle \ y={w(x)v(x)w(x)}}$

${\displaystyle \ y={w^{2}(x)v(x)}}$
${\displaystyle \ y'={2w(x)(w'(x))v(x)+w^{2}v'(x)}}$

BY JUAN CARLOS ARANGO

## PROBLEMAS DE RAZON DE CAMBIO

8)Con que rapídes baja el nivel de un fluido contenido en un tanque cilindrico de almacenamiento si bombeamos hacia fuera el fluido a razon 3000 l/min

${\displaystyle {\frac {dh}{dt}}=3000l/min}$
${\displaystyle \ v={pir^{2}h(metroscubicos)}}$
${\displaystyle \ m^{3}={1000l}}$
${\displaystyle {\frac {dv}{dt}}=1000pir^{2}({\frac {dh}{dt}})}$
${\displaystyle {\frac {-3000}{1000pir^{2}}}={\frac {dh}{dt}}}$
${\displaystyle {\frac {-3000}{1000pir^{2}}}={\frac {dh}{dt}}}$
${\displaystyle {\frac {-3000}{1000pir^{2}}}={\frac {dh}{dt}}}$
${\displaystyle {\frac {-3l}{pir^{2}}}={\frac {dh}{dt}}}$

BY JUAN CARLOS ARANGO

9)un globo de aire caliente se eleva vericalmente desde un campo y es raestreado por un localizador a 500m del punto de despeje. en el instante en que el angulo del localizador es pi/4 el angulo se esta elevando a razon de 0.14 rad/min ¿conque velocidad se esta elevando el globo en ese instante?

${\displaystyle {\frac {dx}{dt}}=0.14rad/min}$
${\displaystyle \ tanx={\frac {h}{500}}}$
${\displaystyle \ h={500}{tanx}}$
${\displaystyle {\frac {dh}{dt}}=500sec^{2}x({\frac {dx}{dt}})}$
${\displaystyle {\frac {dh}{dt}}=500sec^{2}(pi/4)0.14}$
${\displaystyle {\frac {dh}{dt}}=500(2)0.14}$
${\displaystyle {\frac {dh}{dt}}=140m/min}$

BY JUAN CARLOS ARANGO