(a) the domain is\n\\( \\{ x | x ^ { 2 } - 1 \\neq 0 \\} = \\{ x | x \\neq \\pm 1 \\} \\)\n\\( = (…

(a) the domain is\n\\( \\{ x | x ^ { 2 } - 1 \\neq 0 \\} = \\{ x | x \\neq \\pm 1 \\} \\)\n\\( = ( - \\infty, - 1 ), ( - 1,1 ), ( 1, \\infty ) \\) (using interval notation).\n(b) the \\( x - \\) and \\( y \\)-intercepts are both 0.\n(c) since \\( f ( - x ) = f ( x ) \\), the function \\( f \\) is even. the curve is symmetric about the \\( y \\)-axis.\n(d) \\( \\lim _ { x \\rightarrow \\pm \\infty } \\frac { 2 x ^ { 2 } } { x ^ { 2 } - 1 } = \\lim _ { x \\rightarrow \\pm \\infty } \\frac { 2 } { 1 - 1 / x ^ { 2 } } = \\square \\)\ntherefore the line \\( y = 2.2 \\) is a horizontal asymptote.\n- since the denominator is 0 when \\( x = \\pm 1 \\), we compute the following limits:\n\\( \\lim _ { x \\rightarrow 1 ^ { + } } \\frac { 2 x ^ { 2 } } { x ^ { 2 } - 1 } = \\infty \\)\n\\( \\lim _ { x \\rightarrow 1 ^ { - } } \\frac { 2 x ^ { 2 } } { x ^ { 2 } - 1 } = - \\infty \\)\n\\( \\lim _ { x \\rightarrow - 1 ^ { + } } \\frac { 2 x ^ { 2 } } { x ^ { 2 } - 1 } = \\infty \\)\n\\( \\lim _ { x \\rightarrow - 1 ^ { - } } \\frac { 2 x ^ { 2 } } { x ^ { 2 } - 1 } = \\infty \\)

(a) the domain is\n\\( \\{ x | x ^ { 2 } - 1 \\neq 0 \\} = \\{ x | x \\neq \\pm 1 \\} \\)\n\\( = ( - \\infty, - 1 ), ( - 1,1 ), ( 1, \\infty ) \\) (using interval notation).\n(b) the \\( x - \\) and \\( y \\)-intercepts are both 0.\n(c) since \\( f ( - x ) = f ( x ) \\), the function \\( f \\) is even. the curve is symmetric about the \\( y \\)-axis.\n(d) \\( \\lim _ { x \\rightarrow \\pm \\infty } \\frac { 2 x ^ { 2 } } { x ^ { 2 } - 1 } = \\lim _ { x \\rightarrow \\pm \\infty } \\frac { 2 } { 1 - 1 / x ^ { 2 } } = \\square \\)\ntherefore the line \\( y = 2.2 \\) is a horizontal asymptote.\n- since the denominator is 0 when \\( x = \\pm 1 \\), we compute the following limits:\n\\( \\lim _ { x \\rightarrow 1 ^ { + } } \\frac { 2 x ^ { 2 } } { x ^ { 2 } - 1 } = \\infty \\)\n\\( \\lim _ { x \\rightarrow 1 ^ { - } } \\frac { 2 x ^ { 2 } } { x ^ { 2 } - 1 } = - \\infty \\)\n\\( \\lim _ { x \\rightarrow - 1 ^ { + } } \\frac { 2 x ^ { 2 } } { x ^ { 2 } - 1 } = \\infty \\)\n\\( \\lim _ { x \\rightarrow - 1 ^ { - } } \\frac { 2 x ^ { 2 } } { x ^ { 2 } - 1 } = \\infty \\)

Answer

Explanation:

Step1: Calculate the limit

When (x\to\pm\infty), we have (\lim_{x\to\pm\infty}\frac{2}{1 - 1/x^{2}}). As (x\to\pm\infty), (\frac{1}{x^{2}}\to0). So (\lim_{x\to\pm\infty}\frac{2}{1 - 1/x^{2}}=\frac{2}{1-0}=2).

Step2: Analyze the limit as (x\to - 1^{+})

For (x\to - 1^{+}), let (x=-1 + h), where (h\to0^{+}). (x^{2}-1=(x - 1)(x + 1)=(-1 + h-1)(-1 + h + 1)=(-2 + h)h). (\frac{2x^{2}}{x^{2}-1}=\frac{2(-1 + h)^{2}}{(-2 + h)h}=\frac{2(1-2h+h^{2})}{(-2 + h)h}). As (h\to0^{+}), the numerator is positive ((\approx2)) and the denominator is negative ((\approx - 2h)). So (\lim_{x\to - 1^{+}}\frac{2x^{2}}{x^{2}-1}=-\infty).

Answer:

For part (D) first blank: (2); for the limit as (x\to - 1^{+}): (-\infty)