# Límites

1) ${\displaystyle \lim _{x\to 0}{\frac {senx}{2x}}}$

${\displaystyle {\frac {1}{2}}\lim _{x\to 0}{\frac {senx}{x}}}$
${\displaystyle {\frac {1}{2}}.1}$
${\displaystyle {\frac {1}{2}}}$ by:Jorge Montes Jorgemontes

2)${\displaystyle \lim _{x\to -1}{\frac {x^{3}-4x^{2}+x+6}{x+1}}}$

${\displaystyle \lim _{x\to -1}{\frac {(x+1)(x^{2}-5x+6)}{x+1}}}$
${\displaystyle \lim _{x\to -1}\ {x^{2}-5x+6}}$
${\displaystyle \ {(-1)^{2}-5(-1)+6}}$
${\displaystyle \ {1+5+6}}$
${\displaystyle \ {12}}$ by:Jorge Montes Jorgemontes

3)${\displaystyle \lim _{x\to 3}{\frac {x^{4}-18x^{2}+81}{(x-3)^{2}}}}$

${\displaystyle \lim _{x\to 3}{\frac {(x-3)(x^{3}+3x^{2}-9x-27)}{(x-3)^{2}}}}$
${\displaystyle \lim _{x\to 3}{\frac {x^{3}+3x^{2}-9x-27}{x-3}}}$
${\displaystyle \lim _{x\to 3}{\frac {(x-3)(x^{2}+6x+9)}{x-3}}}$
${\displaystyle \lim _{x\to 3}\ {x^{2}+6x+9}}$
${\displaystyle \ {(3)^{2}+6(3)+9}}$
${\displaystyle \ {9+18+9}}$
${\displaystyle \ {36}}$ by:Jorge Montes Jorgemontes

4)${\displaystyle \lim _{h\to 0}{\frac {(x+h)^{2}-x^{2}}{h}}}$

${\displaystyle \lim _{h\to 0}{\frac {x^{2}+2hx+h^{2}-x^{2}}{h}}}$
${\displaystyle \lim _{h\to 0}{\frac {2hx+h^{2}}{h}}}$
${\displaystyle \lim _{h\to 0}{\frac {(h)(2x+h)}{h}}}$
${\displaystyle \lim _{h\to 0}\ {(2x+h)}}$
${\displaystyle \ {(2x)}}$ by:Jorge Montes Jorgemontes

5) ${\displaystyle \lim _{x\to 0}{\frac {tan^{2}x}{senx}}}$

${\displaystyle \lim _{x\to 0}{\frac {(sen^{2}x/cos^{2}x)}{senx}}}$
${\displaystyle \lim _{x\to 0}{\frac {2senxcosx}{(senx)(cos^{2}x-sen^{2}x)}}}$
${\displaystyle 2\lim _{x\to 0}{\frac {cosx}{(cos^{2}x-sen^{2}x)}}}$
${\displaystyle 2\lim _{x\to 0}{\frac {cosx}{cos^{2}x-(1-cos^{2}x)}}}$
${\displaystyle 2\lim _{x\to 0}{\frac {cosx}{2cos^{2}x-1}}}$
${\displaystyle {\frac {2\lim _{x\to 0}\ cosx}{2\lim _{x\to 0}\ 2cos^{2}x-1}}}$
${\displaystyle {\frac {(2)(1)}{(2)(1)-1}}}$
${\displaystyle \ 2}$ by:Jorge Montes Jorgemontes

6)${\displaystyle {\frac {1}{\sqrt[{2}]{3x}}}}$

${\displaystyle \lim _{h\to 0}{\frac {{\frac {1}{\sqrt[{2}]{3x+3h}}}-{\frac {1}{\sqrt[{2}]{3x}}}}{h}}}$
${\displaystyle \lim _{h\to 0}{\frac {\frac {{\sqrt {3x}}-{\sqrt {3x+3h}}}{({\sqrt {3x+3h}})({\sqrt {3x}})}}{h}}}$
${\displaystyle \lim _{h\to 0}{\frac {{\sqrt {3x}}-{\sqrt {3x+3h}}}{(h)(({\sqrt {3x+3h}})({\sqrt {3x}}))}}}$
${\displaystyle \lim _{h\to 0}{\frac {{\sqrt {3x}}-{\sqrt {3x+3h}}}{(h)(({\sqrt {3x+3h}})({\sqrt {3x}}))}}}$
${\displaystyle \lim _{h\to 0}{\frac {{\sqrt {3x}}-{\sqrt {3x+3h}}}{({\sqrt {h^{2}}})(({\sqrt {3x+3h}})({\sqrt {3x}}))}}}$
${\displaystyle \lim _{h\to 0}{\frac {{\sqrt {3x}}-{\sqrt {3x+3h}}}{({\sqrt {(h^{2})(3x^{2}+9hx)}})}}}$
${\displaystyle \lim _{h\to 0}{\frac {{\sqrt {3x}}-{\sqrt {3x+3h}}}{\sqrt {(3x^{2}h^{2}+9h^{3}x)}}}}$
${\displaystyle \lim _{h\to 0}{\frac {{\sqrt {3x}}-{\sqrt {3x+3h}}}{\sqrt {3x^{2}h^{2}+9h^{3}x}}}}$
${\displaystyle \lim _{h\to 0}{\frac {{\sqrt {3x}}-{\sqrt {3x+3h}}}{\sqrt {(3h^{2}x)(3x+3h)}}}}$

1)${\displaystyle \lim _{h\to 0}{\frac {(2+h)^{2}-4}{h}}}$

${\displaystyle \lim _{h\to 0}{\frac {4+4h+h^{2}-4}{h}}}$
${\displaystyle \lim _{h\to 0}{\frac {4h+h^{2}}{h}}}$
${\displaystyle \lim _{h\to 0}{\frac {h(4+h)}{h}}}$
${\displaystyle \lim _{h\to 0}\ {4+h}}$
${\displaystyle \ {4+0}}$
${\displaystyle \ {4}}$ by:Jorge Montes

2)Derivar:${\displaystyle \ F(x)={\frac {(x+1)^{2}}{x-1}}}$

${\displaystyle \lim _{h\to 0}{\frac {{\frac {(((x+h)+1)^{2})}{(x+h)-1}}-{\frac {(x+1)^{2}}{x-1}}}{h}}}$
${\displaystyle \lim _{h\to 0}{\frac {\frac {hx^{2}+h^{2}x-3h-2hx}{(x+h-1)(x-1)}}{h}}}$
${\displaystyle \lim _{h\to 0}{\frac {(h)(x^{2}+hx-3-2x)}{(h)((x+h-1)(x-1))}}}$
${\displaystyle \lim _{h\to 0}{\frac {x^{2}+hx-3-2x}{(x+h-1)(x-1)}}}$
${\displaystyle {\frac {x^{2}-2x-3}{(x-1)^{2}}}}$ by: jorge montes Jorgemontes

y=x
${\displaystyle \ y'=1}$
${\displaystyle \ lim_{x\to 0}{\frac {f(x+h)-F(x)}{h}}}$
${\displaystyle \ lim_{x\to 0}{\frac {x+h-x}{h}}}$
${\displaystyle \ lim_{x\to 0}{\frac {h}{h}}}$
${\displaystyle \ 1}$

by: jorge montes and Zorraidorsito Jorgemontes

4)Derive:${\displaystyle {\frac {6}{x^{2}+1}}}$

${\displaystyle \lim _{h\to 0}{\frac {{\frac {6}{(x+h)^{2}+1}}-{\frac {6}{x^{2}+1}}}{h}}}$
${\displaystyle \lim _{h\to 0}{\frac {\frac {6x^{2}+6-6x^{2}-12hx-6h^{2}-6}{((x+h)^{2}+1)(x^{2}+1)}}{h}}}$
${\displaystyle \lim _{h\to 0}{\frac {\frac {-6h^{2}-12hx}{((x+h)^{2}+1)(x^{2}+1)}}{h}}}$
${\displaystyle \lim _{h\to 0}{\frac {-6h^{2}-12hx}{(h)(((x+h)^{2}+1)(x^{2}+1))}}}$
${\displaystyle \lim _{h\to 0}{\frac {(h)(-6h-12x)}{(h)(((x+h)^{2}+1)(x^{2}+1))}}}$
${\displaystyle \lim _{h\to 0}{\frac {-6h-12x}{((x+h)^{2}+1)(x^{2}+1)}}}$
${\displaystyle {\frac {-6(0)-12x}{((x+(0))^{2}+1)(x^{2}+1)}}}$
${\displaystyle {\frac {-12x}{(x^{2}+1)^{2}}}}$ by: jorge Montes Jorgemontes

5)Derive por implícita: ${\displaystyle \ {y^{2}-x^{2}=1}}$

${\displaystyle \ {{\frac {d}{dx}}y^{2}-{\frac {d}{dx}}x^{2}={\frac {d}{dx}}1}}$
${\displaystyle \ {{\frac {d}{dx}}2y-{\frac {d}{dx}}2x={\frac {d}{dx}}0}}$
${\displaystyle \ {{\frac {d}{dx}}2y-{\frac {d}{dx}}2x=0}}$
${\displaystyle \ {{\frac {d}{dx}}(2y-2x)=0}}$
${\displaystyle \ {{\frac {d}{dx}}(2y)=2x}}$
${\displaystyle \ {{\frac {d}{dx}}={\frac {2x}{2y}}}}$
${\displaystyle \ {{\frac {d}{dx}}={\frac {x}{y}}}}$

6)calcular ${\displaystyle \ f^{,}(0),f^{,}({\frac {1}{2}}),f^{,}(1),f^{,}(-10)}$

${\displaystyle \ {f(x)=2+x-x^{2}}}$
${\displaystyle \ {f^{,}(x)=1-2x}}$
${\displaystyle \ {f^{,}(0)=1}}$
${\displaystyle \ {f^{,}({\frac {1}{2}})=0}}$
${\displaystyle \ {f^{,}(1)=-1}}$
${\displaystyle \ {f^{,}(-10)=-19}}$

# Recta Tangente

Encuentre la ecuación de la recta tangente en el punto propuesto

1)${\displaystyle \ {x^{3}y+y^{3}x=30}}$

# Parcial

1)Se da una tabla de valores para f, g, f´ y

/ x / f(x) / g(x) / f´(x) / g´(x) /
/ 1 / 3 / 2 / 4 / 6 /
/ 2 / 1 / 8 / 5 / 7 /
/ 3 / 7 / 2 / 7 / 9 /
Si h(x) = g(f(x)), el valor de h´(1) es igual a:
a)4 b)5 c)36 d)24 e)30
R/

2)

3)Un rectángulo tiene dos vértices sobre el eje x y otros dos sobre la parabola ${\displaystyle \ {y=12-x^{2}}}$ ¿Cuáles son las dimensiones del rectángulo de este tipo con área máxima?

${\displaystyle \ {A=(2x)(y)}}$
${\displaystyle \ {A=(2x)(12-x^{2})}}$
${\displaystyle \ {A=24x-2x^{3}}}$
${\displaystyle \ {A^{,}=24-6x^{2}}}$
${\displaystyle \ {0=24-6x^{2}}}$
${\displaystyle \ {24=6x^{2}}}$
${\displaystyle \ {{\frac {24}{6}}=x^{2}}}$
${\displaystyle \ {4=x^{2}}}$
${\displaystyle \ {{\sqrt {4}}=x}}$
${\displaystyle \ {2=x}}$
${\displaystyle \ {y=12-x^{2}}}$
${\displaystyle \ {y=12-(2)^{2}}}$
${\displaystyle \ {y=12-4}}$
${\displaystyle \ {y=8}}$
${\displaystyle \ {A=(2(2))(8)}}$
${\displaystyle \ {A=(4)(8)}}$
${\displaystyle \ {A=32}}$

--Jorgemontes 13:07, 19 Jul 2004 (UTC)

4)