Question

Admitindo que x pertence ao domínio da função dada, use as propriedades e os teoremas sobre limite para calcular:

$\mathsf{\lim_{\ x \to -\infty} \sqrt{x^{2}+3x+4}-x}$

Answer 1

Resolução da questão, vejamos:

Tomemos o limite:

$\mathsf{\displaystyle\lim_{x~\to~ -\infty}\sqrt{x^{2}+3x+4}-x}$

Vamos primeiramente dividir e multiplicar o numerador pela função dada, vejamos:

$\mathsf{\displaystyle\lim_{x~\to~ -\infty}\sqrt{x^{2}+3x+4}-x}\\\\\\\\ =\mathsf{\displaystyle\lim_{x~\to~-\infty}~\left(\dfrac{1}{x+\sqrt{x^{2}+3x+4}}(-x+\sqrt{x^{2}+3x+4})(x+\sqrt{x^{2}+3x+4})\right)}}}}\\\\\\\\ =\mathsf{\displaystyle\lim_{x~\to~-\infty}~\left(\dfrac{-x^{2}+(\sqrt{x^{2}+3x+4})^{2}}{x+\sqrt{x^{2}+3x+4}}}}}}\right)}}}}}\\\\\\\\ \mathsf{\displaystyle\lim_{x~\to~-\infty}~\left(\dfrac{3x+4}{x+\sqrt{x^{2}+3x+4}\right)}}}$

Agora dividimos o numerador e o denominador por x, veja:

$\mathsf{\displaystyle\lim_{x~\to~-\infty}~\left(\dfrac{3+\frac{4}{x}}{1+\frac{1}{x}+\sqrt{x^{2}+3x+4}\right)}}}\\\\\\\\ =\mathsf{\displaystyle\lim_{x~\to~-\infty}~\left(\dfrac{3+\frac{4}{x}}{\sqrt{x^{\frac{1}{2}}(x^{2}+3x+4)}+1\right)}}}}\\\\\\\\ =\mathsf{\displaystyle\lim_{x~\to~-\infty}~\left(\dfrac{3+\frac{4}{x}}{\sqrt{1+\frac{3}{x}+\frac{4}{x^{2}}+1\right)}}}}}$

Agora vamos fazer a seguinte substituição:

$\mathsf{u=\dfrac{1}{x}}}}$

Veja:

$\mathsf{\displaystyle\lim_{x~\to~-\infty}~\left(\dfrac{3+\frac{4}{x}}{\sqrt{1+\frac{3}{x}+\frac{4}{x^{2}}+1\right)}}}}}\\\\\\\\=\mathsf{\displaystyle\lim_{u~\to~0^{+}}~\left(\dfrac{4u^{2}+3u}{\sqrt{4u^{2}+3u+1}\right)}}}\\\\\\\\\=\mathsf{1+\dfrac{0~\cdot~4+3}{\sqrt{0~\cdot~3+4~\cdot~0^{2}+1}\right)}=\infty}}}}}}}\\\\\\\\\ \Large\boxed{\boxed{\boxed{\boxed{\mathbf{~\therefore~\displaystyle\lim_{x~\to~ -\infty}~\sqrt{x^{2}+3x+4}-x}=\mathbf{\infty}}}}}}}}}}}}~~\checkmark}}}$

Espero que te ajude. (^.^)