# User:"Jose Freyle"

## LIMITES

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

#### Ejercicio 1

${\displaystyle \lim _{x\to 0}{\frac {sen(x)}{3x}}}$
=${\displaystyle {\frac {1}{3}}\lim _{x\to 0}{\frac {3}{1}}{\frac {sen(x)}{x}}}$
=${\displaystyle {\frac {1}{3}}*1}$
=${\displaystyle {\frac {1}{3}}}$


#### Ejercicio 2

${\displaystyle \lim _{x\to 0}{\frac {sen(x)}{x}}}$
=${\displaystyle \lim _{x\to 0}{\frac {5}{1}}{\frac {sen(5x)}{5x}}}$
=${\displaystyle {\frac {5*1}{1}}}$
=${\displaystyle \ 5(1)}$
=${\displaystyle \ 5}$


#### Ejercicio 3

${\displaystyle \lim _{x\to 0}{\frac {(x^{3}+16x)}{x^{2}+4x}}}$
=${\displaystyle \lim _{x\to 0}{\frac {x(x^{2}+16)}{x(x+4)}}}$
=${\displaystyle \lim _{x\to 0}{\frac {x^{2}+16}{x+4}}}$
=${\displaystyle {\frac {0^{2}+16}{0+4}}}$
=${\displaystyle {\frac {16}{4}}}$
=${\displaystyle \ 4}$


#### Ejercicio 4

${\displaystyle \lim _{x\to 1}{\frac {x^{2}+3x-4}{x-1}}}$
=${\displaystyle \lim _{x\to 1}{\frac {(x-4)(x-1)}{x-1}}}$
=${\displaystyle \ 1+4}$
=${\displaystyle \ 5}$


#### Ejercicio 5

${\displaystyle \lim _{x\to 3}{\frac {2}{x}}+1}$
=${\displaystyle \lim _{x\to 3}{\frac {2+x}{x}}}$
=${\displaystyle {\frac {2+3}{3}}}$
=${\displaystyle {\frac {5}{3}}}$


#### Ejercicio 6

${\displaystyle \lim _{x\to 1}{\frac {5x-x^{2}}{x^{2}+2x-4}}}$
=${\displaystyle {\frac {5(1)-(1)^{2}}{12+2(1)-4}}}$
=${\displaystyle {\frac {5-1}{1+2-4}}}$
=${\displaystyle {\frac {6}{-1}}}$
=${\displaystyle \ -4}$


#### Ejercicio 7

${\displaystyle \lim _{x\to 1}{\frac {x^{2}-x}{2x^{+}5x-7}}}$
=${\displaystyle \lim _{x\to 1}{\frac {x(x-1)}{7(x-1)}}}$
=${\displaystyle {\frac {1}{7}}}$


#### Ejercicio 8

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


#### Ejercicio 9

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


#### Ejercicio 10

 ${\displaystyle \lim _{x\to \infty }{\frac {x^{3}-2x+1}{x^{2}+1}}}$
=${\displaystyle \lim _{x\to \infty }{\frac {{\frac {x^{3}}{x^{2}}}-{\frac {2x}{x^{2}}}+{\frac {1}{x^{2}}}}{{\frac {x^{2}}{x^{2}}}+{\frac {1}{x^{2}}}}}}$
=${\displaystyle {\frac {x-0+0}{1+0}}}$
=${\displaystyle {\frac {x}{1}}}$
=${\displaystyle \infty }$


#### Ejercicio 11

${\displaystyle \lim _{x\to \infty }{\frac {5}{x^{2}+8x+15}}}$
=${\displaystyle \lim _{x\to \infty }{\frac {\frac {5}{x^{2}}}{{\frac {x^{2}}{x^{2}}}+{\frac {8x}{x^{2}}}+{\frac {15}{x^{2}}}}}}$
=${\displaystyle {\frac {0}{1+0+0}}}$
=${\displaystyle {\frac {0}{1}}}$
=${\displaystyle \ 0}$


#### Ejercicio 12

${\displaystyle \lim _{x\to \infty }{\frac {1}{x^{2}+5x-6}}}$
=${\displaystyle \lim _{x\to \infty }{\frac {\frac {1}{x^{2}}}{{\frac {x^{2}}{x^{2}}}+{\frac {5x}{x^{2}}}-{\frac {6}{x^{2}}}}}}$
=${\displaystyle {\frac {0}{1+0-0}}}$
=${\displaystyle {\frac {0}{1}}}$
=${\displaystyle \ 0}$


#### Ejercicio 13

${\displaystyle \lim _{t\ 0}{\frac {(}{\sqrt[{2}]{2t}}})-{\sqrt[{2}]{2}}){t}}$
${\displaystyle \lim _{t\ 0}{\frac {{\sqrt[{2}]{2t}}-{\sqrt[{2}]{2}}}{t}}*{\sqrt[{2}]{2t}}+{\sqrt[{2}]{2}}{\sqrt[{2}]{2t}}+{\sqrt[{2}]{\sqrt[{2}]{2}}}}$
${\displaystyle \lim _{t\ 0}{\frac {-t}{t({\sqrt[{2}]{2t}}+{\sqrt[{2}]{)}}}}}$
${\displaystyle \lim _{t\ 0}{\frac {-1}{({\sqrt[{2}]{2t}}+{\sqrt[{2}]{)}}}}}$
${\displaystyle {\frac {-1}{({\sqrt[{2}]{2*0}}+{\sqrt[{2}]{)}}}}}$
${\displaystyle {\frac {-1}{({0}+{\sqrt[{2}]{2}})}}}$
${\displaystyle {\frac {-1}{({\sqrt[{2}]{2}})}}}$


#### Ejercicio 14

${\displaystyle \lim _{x\to \infty }\ ((x^{5}+7x^{4}+2)^{(}c)-x)}$
${\displaystyle \lim _{x\to \infty }{\frac {(x^{5}+7x^{4}+2)^{(}2c)}{(x^{5}+7x^{4}+2)^{(}c)+x}}}$


${\displaystyle \lim _{x\to \infty }\ ((x^{5}+7x^{4}+2)^{(}{\frac {1}{5}})-x)}$

${\displaystyle {\frac {7}{5}}}$


#### Ejercicio 1

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



#### Ejercicio 2

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


#### Ejercicio 3

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


#### Ejercicio 4

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


#### Ejercicio 5

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