User:"Jose Freyle"

From Wikibooks, open books for an open world
Jump to: navigation, search

LIMITES[edit]

En esta sección encontraremos la solución de algunos problemas relacionados con limites al infinito, trigonométricos y algunos que se demuestran directamente.

Ejercicio 1[edit]

\lim_{x\to 0}\frac {sen(x)}{3x}
 =\frac {1}{3}\lim_{x\to 0}\frac {3}{1}\frac {sen(x)}{x}
 =\frac {1}{3} * 1
 =\frac {1}{3}

Ejercicio 2[edit]

\lim_{x\to 0}\frac {sen(x)}{x}
 =\lim_{x\to 0}\frac {5}{1}\frac {sen(5x)}{5x}
 =\frac {5 * 1}{1}
 =\ 5 (1)
 =\ 5

Ejercicio 3[edit]

\lim_{x\to 0}\frac {(x^3+16x)}{x^2+4x}
 =\lim_{x\to 0}\frac {x(x^2+16)}{x(x+4)}
 =\lim_{x\to 0}\frac {x^2+16}{x+4}
 =\frac {0^2+16}{0+4}
 =\frac {16}{4}
 =\ 4

Ejercicio 4[edit]

\lim_{x\to 1}\frac {x^2+3x-4}{x-1}
 =\lim_{x\to 1}\frac {(x-4)(x-1)}{x-1}
 =\ 1 + 4 
 =\ 5

Ejercicio 5[edit]

\lim_{x\to 3}\frac {2}{x}+1
 =\lim_{x\to 3}\frac {2 + x}{x}
 =\frac {2+3}{3}
 =\frac {5}{3} 

Ejercicio 6[edit]

\lim_{x\to 1}\frac {5x-x^2}{x^2+2x-4}
 =\frac {5(1)-(1)^2}{12+2(1)-4}
 =\frac {5-1}{1+2-4}
 =\frac {6}{-1}
 =\ -4

Ejercicio 7[edit]

\lim_{x\to 1}\frac {x^2-x}{2x^+5x-7}
 =\lim_{x\to 1}\frac {x(x-1)}{7(x-1)}
 =\frac {1}{7}

Ejercicio 8[edit]

\lim_{x\to -1}\frac {x^2-1}{x^2+3x+2}
 =\lim_{x\to -1}\frac {(x+1)(x-1)}{(x+2)(x+1)}
 =\frac {-1-1}{-1+2}
 =\frac {-2}{1}
 =\ -2 

Ejercicio 9[edit]

\lim_{x\to 2}\frac {3x^5-8x^4+5x^3-3x-2}{7x^2-6x-16}
=\lim_{x\to 2}\frac {(x-2)(3x^4-2x^3+x^2+2x+1)}{(x-2)(7x+8)}
=\frac {48-16+4+4+1}{14+8}
=\frac {41}{22}

Ejercicio 10[edit]

 \lim_{x\to \infty}\frac {x^3-2x+1}{x^2+1}
=\lim_{x\to \infty}\frac {\frac{x^3}{x^2}-\frac{2x}{x^2}+\frac{1}{x^2}}{\frac{x^2}{x^2}+\frac{1}{x^2}}
=\frac {x-0+0}{1+0}
=\frac {x}{1}
=\infty 

Ejercicio 11[edit]

\lim_{x\to \infty}\frac {5}{x^2+8x+15}
=\lim_{x\to \infty}\frac {\frac {5}{x^2}}{\frac{x^2}{x^2}+ \frac {8x}{x^2}+\frac {15}{x^2}}
=\frac{0}{1+0+0}
=\frac {0}{1}
=\ 0

Ejercicio 12[edit]

\lim_{x\to \infty}\frac{1}{x^2+5x-6}
=\lim_{x\to \infty}\frac{\frac{1}{x^2}}{\frac{x^2}{x^2}+\frac{5x}{x^2}-\frac{6}{x^2}}
=\frac{0}{1+0-0}
=\frac{0}{1}
=\ 0


Ejercicio 13[edit]

\lim_{t\ 0}\frac (\sqrt[2]{2t})-\sqrt[2]{2}){t}
\lim_{t\ 0}\frac {\sqrt[2]{2t}-\sqrt[2]{2}}{t}* \sqrt[2]{2t}+\sqrt[2]{2} \sqrt[2]{2t}+\sqrt[2]\sqrt[2]{2}
\lim_{t\ 0}\frac {-t}{t(\sqrt[2]{2t}+\sqrt[2])}
\lim_{t\ 0}\frac {-1}{(\sqrt[2]{2t}+\sqrt[2])}
\frac {-1}{(\sqrt[2]{2*0}+\sqrt[2])}
\frac {-1}{({0}+\sqrt[2]{2})}
\frac {-1}{(\sqrt[2]{2})}

Ejercicio 14[edit]

\lim_{x\to \infty}\ ((x^5+7x^4+2)^(c)-x)
\lim_{x\to \infty}\frac {(x^5+7x^4+2)^(2c)}{(x^5+7x^4+2)^(c)+x}


\lim_{x\to \infty}\ ((x^5+7x^4+2)^(\frac {1}{5})-x)
\frac {7}{5}

DERIVADAS[edit]

Ejercicio 1[edit]

\ y=(3x^5-1)(2x+3)
\ y'= (3x^5-1)(2)+(2x+3)(15x^4)
\ y'= 6x^5-2+30x^5+45x^4
\ y'= 36x^5+45x^4-2

Ejercicio 2[edit]

\ y=6x^4+3x^3+6x^2-8x+2
\ y'=24x^3+9x^2+12x-8

Ejercicio 3[edit]

\ f(x)= cos(x)
f'(x)= lim_{h\to 0}\frac {f(x+h)-f(x)}{h}
f'(x)= lim_{h\to 0}\frac {cos(x+h)-cosx}{h}
\lim_{h\to 0}\frac {cos(x)*cos(h)-sen(x)*sen(h)-cos(x)}{h}
\lim_{h\to 0}\frac {cos(x)(cos h-1)}{h}* \frac {sen(x)sen(h)}{h}
\cos(x)* lim_{h\to 0}\frac {cos(h)-1}{h}-sen(lim_{h\to 0}\frac {sen(h)}{h}
\lim_{h\to 0}\ 0*cos(x)- lim_{h\to 0}\ 1* -sen(x)
\ 0-sen(x)
\ - sen(x)

Ejercicio 4[edit]

\ f(x)= x*e^{-x}
\ f'(x)= e^{-x}(1-x)

Ejercicio 5[edit]

\ f(x)=5x^5+4x^4+3x^3+2x^2+x
\ f'(x)= 25x^4+16x^3+9x^2+4x+1