Question

determine caso exista o limite (x+10) /in (x2+1) quando tende a 0

Answer 1

Resposta:

$\lim_{x \to \00} \frac{x+10}{ln(x^2+1)} =\infty$

Explicação passo a passo:

$\lim_{x \to \0 0} \frac{x+10}{ln(x^2+1)} =\frac{0+10}{ln(0^2+1)} =\frac{10}{ln1} =\frac{10}{0}~(nada~podemos~concluir.)$

Calculado os limites laterais:

$\lim_{ \to \00_+} \frac{x+10}{ln(x^2+1)} = \lim_{ \to \00_+} \frac{0+10}{ln(0_+^2+1)} =\frac{10}{ln1} =\frac{10}{0_+}= \infty$

$\lim_{ \to \00_} \frac{x+10}{ln(x^2+1)} = \lim_{ \to \00_-} \frac{0+10}{ln(0_-^2+1)} =\frac{10}{ln(0+1)} =\frac{10}{ln1}=\frac{10}{0_+} = \infty}$

$Logo,~ \lim_{x \to \00} \frac{x+10}{ln(x^2+1)} =\infty$