Límites indeterminados 0/0

Límites indeterminados 0/0

12th Grade

12 Qs

quiz-placeholder

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Factorizing #1

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Límites indeterminados 0/0

Límites indeterminados 0/0

Assessment

Quiz

Mathematics

12th Grade

Hard

Created by

ROMÁN CLEMENTE

Used 43+ times

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12 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

10 sec • 1 pt

¿Qué condición se debe cumplir para que un límite sea considerado indeterminado?

0\frac{0}{\infty}

0\frac{\infty}{0}

00\frac{0}{0}

\infty

2.

MULTIPLE CHOICE QUESTION

10 sec • 1 pt

¿Qué procedimiento usarías para poder resolver el siguiente límite indeterminado?

 limx2 x24x2\lim_{x\rightarrow2}\ \frac{x^2-4}{x-2}  

Sustitución directa

Factorización

Racionalización

No se puede resolver

3.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

¿Cuál es el resultado al resolver el limite?

 limx2 x24x2\lim_{x\rightarrow2}\ \frac{x^2-4}{x-2}  

 14\frac{1}{4}  

 14-\frac{1}{4}  

4

 00\frac{0}{0}  

4.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

Resolviendo el límite siguiente nos da como resultado:

 limx4 x2x4\lim_{x\rightarrow4}\ \frac{\sqrt{x}-2}{x-4}  

 00\frac{0}{0}  

 14\frac{1}{4}  

 44  

 14-\frac{1}{4}  

5.

MULTIPLE CHOICE QUESTION

20 sec • 1 pt

El conjugado de la expresión

 x43\sqrt{x-4}-3  es:

 x4+3\sqrt{x-4}+3  

 x+43\sqrt{x+4}-3  

 x+4+3\sqrt{x+4}+3  

 x4+3-\sqrt{x-4}+3  

6.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

Resolviendo el limite que a continuación se presenta, nos queda:

 limx5 x22x15x5\lim_{x\rightarrow5}\ \frac{x^2-2x-15}{x-5}  

 00\frac{0}{0}  

 15\frac{1}{5}  

 55  

 15-\frac{1}{5}  

7.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

El siguiente límite es indeterminado, si factorizas las expresiones, te quedaría como:

 limx1 x21x2+3x+2\lim_{x\rightarrow-1}\ \frac{x^2-1}{x^2+3x+2}  

 (x+1)(x1)(x+2)(x+1)\frac{\left(x+1\right)\left(x-1\right)}{\left(x+2\right)\left(x+1\right)}  

 (x1)(x1)(x+2)(x+1)\frac{\left(x-1\right)\left(x-1\right)}{\left(x+2\right)\left(x+1\right)}  

 (x+1)(x1)(x+2)(x1)\frac{\left(x+1\right)\left(x-1\right)}{\left(x+2\right)\left(x-1\right)}  

 (x+1)(x1)(x2)(x+1)\frac{\left(x+1\right)\left(x-1\right)}{\left(x-2\right)\left(x+1\right)}   

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