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Question: \(\infty)f(x) = \log_{3 + x}(x^{2} - 1)\) equals to –...

∞)f(x)=log⁡3+x(x2−1)\infty)f(x) = \log_{3 + x}(x^{2} - 1) equals to –

A

– (−3,−1)∪(1,∞)( - 3, - 1) \cup (1,\infty)

B

[−3,−1)∪[1,∞)\lbrack - 3, - 1) \cup \lbrack 1,\infty)

C

–(−3,−2)∪(−2,−1)∪(1,∞)( - 3, - 2) \cup ( - 2, - 1) \cup (1,\infty)

D

[−3,−2)∪(−2,−1)∪[1,∞]\lbrack - 3, - 2) \cup ( - 2, - 1) \cup \lbrack 1,\infty\rbrack

Answer

– (−3,−1)∪(1,∞)( - 3, - 1) \cup (1,\infty)

Explanation

Solution

lim⁡x→2\lim _ { x \rightarrow 2 } (4x2−4−1x−2)\left( \frac { 4 } { x ^ { 2 } - 4 } - \frac { 1 } { x - 2 } \right) = lim⁡x→2\lim _ { x \rightarrow 2 } (4−x−2x2−4)\left( \frac { 4 - x - 2 } { x ^ { 2 } - 4 } \right)

=lim⁡x→2\lim _ { x \rightarrow 2 } −(x−2)(x+2)(x−2)\frac { - ( x - 2 ) } { ( x + 2 ) ( x - 2 ) } = −14\frac { - 1 } { 4 }