Question

resolva o limite
$\lim_{x \to \-1/2} \frac{x^{2}+ x+ 2}{2x^{2} +5x+2}$

Answer 1

Explicação passo-a-passo:

Cálculo do Limite:

$\mathsf{\[\lim_{x~\rightarrow~-\dfrac{1}{2}} \dfrac{\mathsf{x^2+x+2}}{\mathsf{2x^2+5x+2}} \] }$

$\mathsf{=\dfrac{\Big(-\dfrac{1}{2}\Big)^2-\dfrac{1}{2}+2}{2.\Big(-\dfrac{1}{2}\Big)^2+5.\Big(-\dfrac{1}{2}\Big)+2} }$

$\mathsf{=~\dfrac{\dfrac{1}{4}+\dfrac{3}{2}}{\dfrac{1}{2}-\dfrac{1}{2} } }$

$\mathsf{=~\dfrac{\dfrac{1}{4}+\dfrac{6}{4}}{0} }$

$\mathsf{=~\dfrac{\dfrac{7}{4}}{0} =~+\infty }$

Espero ter ajudado bastante!)

Answer 2

$\lim_{x \to \ - \frac{1}{2} } \frac{ {x}^{2} + x + 2}{2 {x}^{2} + 5x + 2} \\ \frac{ {( - \frac{1}{2} )}^{2} - \frac{1}{2} + 2}{2. {( - \frac{1}{2} )}^{2} + 5( - \frac{1}{2}) + 2 }$

$\frac{ \frac{1}{4} - \frac{1}{2} + 2}{2.( \frac{1}{4} ) - \frac{5}{2} + 2} \\ = \frac{ \frac{1 - 2 + 8}{4} }{ \frac{2 - 10 + 8}{4} }$

$\frac{ \frac{7}{4} }{0} = \infty$