Encontre o limite, se existir.
.
Passo 1
Estamos calculando o limite de \
quando \
.
Passo 2
Para lidar com a raiz, fatoramos $x^{2}$ dentro da raiz:
$$
\sqrt{x{2}-9}=\sqrt{x{2}\bigl(1-\tfrac{9}{x{2}}\bigr)}=|x|\sqrt{1-\tfrac{9}{x{2}}}.
$$
\sqrt{x{2}-9}=\sqrt{x{2}\bigl(1-\tfrac{9}{x{2}}\bigr)}=|x|\sqrt{1-\tfrac{9}{x{2}}}.
$$
Como $x\to -\infty$, temos $|x|=-x$, logo $\sqrt{x{2}-9}=-x\sqrt{1-\tfrac{9}{x{2}}}.$
Passo 3
Substituímos essa expressão no quociente:
$$
\frac{\sqrt{x^{2}-9}}{2x-6}=
\frac{-x\sqrt{1-\tfrac{9}{x^{2}}}}{2x\bigl(1-\tfrac{3}{x}\bigr)}.
$$
\frac{\sqrt{x^{2}-9}}{2x-6}=
\frac{-x\sqrt{1-\tfrac{9}{x^{2}}}}{2x\bigl(1-\tfrac{3}{x}\bigr)}.
$$
Agora fatoramos $x$ em cima e embaixo para simplificar:
$$
\frac{-x}{2x}\cdot\frac{\sqrt{1-\tfrac{9}{x{2}}}}{1-\tfrac{3}{x}}=-\frac{1}{2}\cdot\frac{\sqrt{1-\tfrac{9}{x{2}}}}{1-\tfrac{3}{x}}.
$$
\frac{-x}{2x}\cdot\frac{\sqrt{1-\tfrac{9}{x{2}}}}{1-\tfrac{3}{x}}=-\frac{1}{2}\cdot\frac{\sqrt{1-\tfrac{9}{x{2}}}}{1-\tfrac{3}{x}}.
$$
Passo 4
Ao tomar o limite $x\to -\infty$, $\tfrac{9}{x^{2}}\to 0$ e $\tfrac{3}{x}\to 0$. Portanto
$$
\lim_{x\to -\infty}\frac{\sqrt{1-\tfrac{9}{x^{2}}}}{1-\tfrac{3}{x}}=\frac{1}{1}=1.
$$
\lim_{x\to -\infty}\frac{\sqrt{1-\tfrac{9}{x^{2}}}}{1-\tfrac{3}{x}}=\frac{1}{1}=1.
$$
Multiplicando por $-\tfrac{1}{2}$ dá $-\tfrac{1}{2}$.
Resposta
-\tfrac{1}{2}