Question

Calcula los siguientes límites ayudaaaaaan pliiiis:(​

Answer 1

Respuesta y pasos:

1) $\lim _{x\to \:2}\left(\frac{x^2+x-6}{x-2}\right)$

$\mathrm{Simplificar}\:\frac{x^2+x-6}{x-2}$

$\frac{x^2+x-6}{x-2}$

$\mathrm{Factorizar}\:x^2+x-6:\quad \left(x-2\right)\left(x+3\right)$

$=\frac{\left(x-2\right)\left(x+3\right)}{x-2}$

$\mathrm{Eliminar\:los\:terminos\:comunes:}\:x-2$

$=x+3$

$=\lim _{x\to \:2}\left(x+3\right)$

$\mathrm{Sustituir\:la\:variable}$

$=2+3$

$\mathrm{Simplificar}$

$=5$

2) $\lim _{x\to -4}\left(\frac{x^2+5x+4}{x^2+3x-4}\right)$

$\mathrm{Simplificar}\:\frac{x^2+5x+4}{x^2+3x-4}$

$\mathrm{Factorizar}\:x^2+5x+4:\quad \left(x+1\right)\left(x+4\right)$

$=\frac{\left(x+1\right)\left(x+4\right)}{x^2+3x-4}$

$\mathrm{Factorizar}\:x^2+3x-4:\quad \left(x-1\right)\left(x+4\right)$

$=\frac{\left(x+1\right)\left(x+4\right)}{\left(x-1\right)\left(x+4\right)}$

$\mathrm{Eliminar\:los\:terminos\:comunes:}\:x+4$

$=\frac{x+1}{x-1}$

$=\lim _{x\to \:-4}\left(\frac{x+1}{x-1}\right)$

$\mathrm{Sustituir\:la\:variable}$

$=\frac{-4+1}{-4-1}$

$\mathrm{Simplificar}$

$=\frac{3}{5}$

3) $\lim _{x\to \:6}\left(\frac{x^2-36}{x-6}\right)$

$\mathrm{Simplificar}\:\frac{x^2-36}{x-6}$

$\mathrm{Factorizar}\:x^2-36:\quad \left(x+6\right)\left(x-6\right)$

$=\frac{\left(x+6\right)\left(x-6\right)}{x-6}$

$\mathrm{Eliminar\:los\:terminos\:comunes:}\:x-6$

$=x+6$

$=\lim _{x\to \:6}\left(x+6\right)$

$\mathrm{Sustituir\:la\:variable}$

$=6+6$

$\mathrm{Simplificar}$

$=12$

4) $\lim _{x\to \:5}\left(\frac{x^3-25x}{x^2-5x}\right)$

$\mathrm{Simplificar}\:\frac{x^3-25x}{x^2-5x}$

$\mathrm{Factorizar}\:x^3-25x:\quad x\left(x^2-25\right)$

$=\frac{x\left(x^2-25\right)}{x^2-5x}$

$\mathrm{Factorizar}\:x^2-5x:\quad x\left(x-5\right)$

$=\frac{x^2-25}{x-5}$

$\mathrm{Factorizar}\:x^2-25:\quad \left(x+5\right)\left(x-5\right)$

$=\frac{\left(x+5\right)\left(x-5\right)}{x-5}$

$\mathrm{Eliminar\:los\:terminos\:comunes:}\:x-5$

$=x+5$

$=\lim _{x\to \:5}\left(x+5\right)$

$Sustituir\:la\:variable$

$=5+5$

$\mathrm{Simplificar}$

$=10$

5) $\lim _{x\to -4}\left(\frac{x^2+3x-4}{x^2+8x+16}\right)$

$\mathrm{Si\:}\lim _{x\to a^-}f\left(x\right)\ne \lim _{x\to a^+}f\left(x\right)\mathrm{\:entonces\:el\:limite\:no\:existe}$

$\lim _{x\to \:-4+}\left(\frac{x^2+3x-4}{x^2+8x+16}\right)=-\infty \:$

$\lim _{x\to \:-4-}\left(\frac{x^2+3x-4}{x^2+8x+16}\right)=\infty \:$

$=\mathrm{Es\:divergente}$