# Transwiki:Calculo diferencial solucionario:Limites algebraicos

Utilizando las propiedades básicas para el cálculo de los limites resuelva los siguientes límites:

 7x-3+2x+1


### Ejercicio 2

${\displaystyle \lim _{x\to 1}{\frac {x^{4}-x^{5}}{1-x}}}$

Respuesta:
${\displaystyle \lim _{x\to 1}{\frac {x^{4}(1-x)}{1-x}}}$
${\displaystyle \lim _{x\to 1}x^{4}}$
${\displaystyle 1}$

\lim\limits_{x\to \frac{1}{2}}\frac{2x^{2}+x-1}{2x^{2}-3x+1}

${\displaystyle \lim _{x\to 2}{\frac {2-{\sqrt[{2}]{x+2}}}{x-2}}\cdot {\frac {2+{\sqrt[{2}]{x+2}}}{2+{\sqrt[{2}]{x+2}}}}}$
${\displaystyle \lim _{x\to 2}{\frac {-(x-2)}{(x-2)(2+{\sqrt[{2}]{x+2}}}}}$
${\displaystyle \lim _{x\to 2}{\frac {-1}{2+{\sqrt[{2}]{x+2}}}}}$

### Ejercicio 27

${\displaystyle \lim _{x\to 0}{\frac {{\sqrt {(}}2-t)-{\sqrt {2}}}{t}}}$

RESPUESTA:
evaluando:

${\displaystyle \lim _{x\to 0}{\frac {{\sqrt {(}}2-t)-{\sqrt {2}}}{t}}}$ =${\displaystyle {\frac {{\sqrt {(}}2-0)-{\sqrt {2}}}{0}}}$ =${\displaystyle {\frac {{\sqrt {2}}-{\sqrt {2}}}{0}}={\frac {0}{0}}}$

${\displaystyle \lim _{x\to 0}{\frac {{\sqrt {(}}2-t)-{\sqrt {2}}}{t}}}$ =${\displaystyle \lim _{x\to 0}{\frac {({\sqrt {(}}2-t)-{\sqrt {2}})({\sqrt {(}}2-t)+{\sqrt {2}}}{t({\sqrt {(}}2-t)+{\sqrt {2}}}}}$ =${\displaystyle \lim _{x\to 0}{\frac {({\sqrt {(}}2-t))^{2}-({\sqrt {2}})^{2}}{t({\sqrt {(}}2-t)+{\sqrt {2}})}}}$ =${\displaystyle \lim _{x\to 0}{\frac {(2-t)-2)}{t({\sqrt {(}}2-t)+{\sqrt {2}})}}}$ =${\displaystyle \lim _{x\to 0}{\frac {-t}{t({\sqrt {(}}2-t)+{\sqrt {2}})}}}$ =${\displaystyle \lim _{x\to 0}{\frac {-1}{{\sqrt {(}}2-t)+{\sqrt {2}}}}={\frac {-1}{{\sqrt {(}}2-0)+{\sqrt {2}}}}}$ =${\displaystyle {\frac {-1}{{\sqrt {(}}2)+{\sqrt {2}}}}}$ =${\displaystyle {\frac {-1}{2{\sqrt {2}}}}}$

### Ejercicio 28

${\displaystyle \lim _{x\to 0}{\frac {x}{{\sqrt {(}}1+3x)-1}}}$

RESPUESTA:
evaluando:

${\displaystyle \lim _{x\to 0}{\frac {x}{{\sqrt {(}}1+3x)-1}}}$ =${\displaystyle {\frac {0}{{\sqrt {(}}1+3(0))-1}}}$ =${\displaystyle {\frac {0}{{\sqrt {1}}-1}}}$ =${\displaystyle {\frac {0}{0}}}$

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

### Ejercicio 30

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

### Ejercicio 32

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

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

### Ejercicio 33

${\displaystyle \lim _{x\to a}{\frac {x^{2}-a^{2}}{x^{2}-2ax+a^{2}}}}$

Respuesta:
${\displaystyle \lim _{x\to a}{\frac {(x+a)(x-a)}{(x-a)(x-a)}}}$
${\displaystyle \lim _{x\to a}{\frac {(x+a)}{(x-a)}}}$

### Ejercicio 34

${\displaystyle lim_{x\to 4}{\frac {x-4}{x^{2}-x-12}}}$

Respuesta:
${\displaystyle lim_{x\to 4}{\frac {x-4}{(x-4)(x+3)}}}$
${\displaystyle lim_{x\to 4}{\frac {1}{x+3}}}$
${\displaystyle 1/7}$

### Ejercicio 35

${\displaystyle lim_{x\to 3}{\frac {x^{3}-27}{x^{2}-9}}}$

Respuesta:
${\displaystyle lim_{x\to 3}{\frac {(x-3)(x^{2}+3x+9)}{(x-3)(x+3)}}}$
${\displaystyle lim_{x\to 3}{\frac {:}{<}}math>lim_{x\to 3}{\frac {x^{2}+3x+9}{x+3}}}$
${\displaystyle lim_{x\to 3}{\frac {9+9+9}{6}}}$
${\displaystyle 27/6}$

${\displaystyle 9/2}$

### Ejercicio 36

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

Respuesta:
${\displaystyle lim_{h\to 0}{\frac {x^{2}+2hx+h^{2}-x^{2}}{h}}}$
${\displaystyle lim_{h\to 0}{2x+h}}$
${\displaystyle 2x}$

### Ejercicio 37

${\displaystyle lim_{x\to \infty }{\frac {3x-2}{8x+7}}}$

Respuesta:

${\displaystyle lim_{x\to \infty }{\frac {3-(2/x)}{8+(7/x)}}}$ ${\displaystyle 3/8}$

### Ejercicio 38

${\displaystyle lim_{x\to \infty }{\frac {6x^{2}+2x+1}{5x^{2}-3x-4}}}$

Respuesta:

${\displaystyle lim_{x\to \infty }{\frac {6+(2/x)+(1/x^{2})}{5-(3/x)-(4/x^{2})}}}$ ${\displaystyle 6/5}$

### Ejercicio 39

${\displaystyle lim_{x\to \infty }{\frac {x^{2}+x-2}{4x^{3}-1}}}$

Respuesta:

${\displaystyle lim_{x\to \infty }{\frac {(1/x)+(1/x^{2})-(2/x^{3})}{4-(1/x^{3})}}}$ ${\displaystyle 0}$

### Ejercicio 40

${\displaystyle lim_{x\to \infty }{\frac {2x^{3}}{x^{2}+1}}}$

Respuesta:

${\displaystyle lim_{x\to \infty }{\frac {2}{(1/x)+(1/x^{3})}}}$ ${\displaystyle n.e}$

### Ejercicio 41

${\displaystyle \lim _{x\to 4}f(x)=\lim _{x\to 4}{\frac {{\sqrt[{2}]{x}}-2}{x-4}}}$
${\displaystyle \lim _{x\to 4}f(x)=\lim _{x\to 4}{\frac {({\sqrt[{2}]{x}}-2)({\sqrt[{2}]{x}}+2)}{(x-4)({\sqrt[{2}]{x}}+2)}}}$
${\displaystyle \lim _{x\to 4}f(x)=\lim _{x\to 4}{\frac {x-4}{(x-4)({\sqrt[{2}]{x}}+2)}}}$ :${\displaystyle \lim _{x\to 4}f(x)=\lim _{x\to 4}{\frac {1}{{\sqrt[{2}]{x}}+2)}}}$
${\displaystyle \lim _{x\to 4}f(x)={\frac {1}{4}}}$

===Ejercicio

### Ejercicio 48

${\displaystyle \lim _{x\to 5}{\frac {1-{\sqrt[{2}]{x-4}}}{x-5}}{\frac {1+{\sqrt[{2}]{x-4}}}{1+{\sqrt[{2}]{x-4}}}}}$

Respuesta:
${\displaystyle \lim _{x\to 5}{\frac {-(x-5)}{(x-5)(1+{\sqrt[{2}]{x+4}})}}}$
${\displaystyle \lim _{x\to 5}{\frac {-1}{1+{\sqrt {x+4}}}}}$
${\displaystyle {\frac {-1}{1+{\sqrt {9}}}}}$
${\displaystyle {\frac {-1}{4}}}$

### Ejercicio 49

49)${\displaystyle \lim _{x\to 6}{\frac {{\sqrt[{2}]{x-2}}-2}{x-6}}{\frac {{\sqrt[{2}]{x-2}}+2}{{\sqrt[{2}]{x-2}}+2}}}$

Respuesta:
${\displaystyle \ lim_{x\to 6}{\frac {x-6}{(x-6)({\sqrt[{2}]{x-2}}+2)}}}$
${\displaystyle \ lim_{x\to 6}{\frac {1}{({\sqrt[{2}]{x-2}}+2)}}}$
${\displaystyle {\frac {1}{{\sqrt {4}}+2}}}$
${\displaystyle {\frac {1}{4}}}$

### Ejercicio 50

50)${\displaystyle \lim _{h\to 0}{\frac {(x+h)^{3}-x^{3}}{h}}}$

Respuesta:
${\displaystyle \lim _{h\to 0}{\frac {x^{3}+3x^{2}h+3xh^{2}+h^{3}-x^{3}}{h}}}$
${\displaystyle \lim _{h\to 0}{\frac {h(3x^{2}+3xh+h^{2})}{h}}}$
${\displaystyle \lim _{h\to 0}3x^{2}+\lim _{x\to 0}3xh+\lim _{h\to 0}h^{2}}$
${\displaystyle 3x^{2}}$

### Ejercicio 51

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

Respuesta:
${\displaystyle \lim _{x\to 2}{\frac {(4-x^{2})(3+{\sqrt {x^{2}+5}})}{4-x^{2}}}}$
${\displaystyle \lim _{x\to 2}3+{\sqrt {(}}x^{2}+5)}$
${\displaystyle 3+3}$
${\displaystyle 6}$

### Ejercicio 52

52)${\displaystyle \lim _{h\to 0}{\frac {{\sqrt[{2}]{x+h}}-{\sqrt[{2}]{x}}}{h}}}$

Respuesta:

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