Question

Alguien me puede ayudar en factorizar estos límites indeterminados por favor!.

Answer 1

4)$\lim_{x \to \frac{3}{2}}\sqrt{\frac{2x-3}{4x^2-12x+9}}\\\\= \lim_{x \to \frac{3}{2}}\sqrt{\frac{2x-3}{(2x-3)^2}}\\\\=\lim_{x \to \frac{3}{2}}\sqrt{\frac{1}{(2x-3)}}=\sqrt{\frac{1}{0}}\ ;\ \text{el limite es infinito}$

5)$\lim_{x \to \frac{3}{2}}\frac{5-2x}{25-4x^2}\\\\=\lim_{x \to \frac{3}{2}}\frac{5-2x}{(5+2x)(5-2x)}\\\\=\lim_{x \to \frac{3}{2}}\frac{1}{5+2x}\\\\=\frac{1}{5+2(\frac{3}{2})}\\\\=\frac{1}{8}$

6)$\lim_{x \to \frac{1}{3}}\frac{3x-1}{3x^2+5x-2}\\\\=\lim_{x \to \frac{1}{3}}\frac{3x-1}{(x+2)(3x-1)}\\\\=\lim_{x \to \frac{1}{3}}\frac{1}{x+2}\\\\=\frac{1}{\frac{1}{3}+2}\\\\=\frac{1}{\frac{7}{3}}\\\\=\frac{3}{7}$

7)$\lim_{x \to -2} \, \frac{2+x}{5x^2+14x+8}\\\\=\lim_{x \to -2} \, \frac{2+x}{(2+x)(5x+4)}\\\\=\lim_{x \to -2} \, \frac{1}{5x+4}\\\\=\frac{1}{5(-2)+4}\\\\=-\frac{1}{6}$

8)$\lim_{x \to -2} \, \frac{3x^2+5x-2}{x+2}\\\\=\lim_{x \to -2} \, \frac{(x+2)(3x-1)}{x+2}\\\\=\lim_{x \to -2} \, 3x-1\\\\=3(-2)-1\\\\=-7$