Answer
$\lim_{x\to1}\Big(\dfrac{x^{4}+x^{2}-6}{x^{4}+2x+3}\Big)^{2}=\dfrac{4}{9}$
Work Step by Step
$\lim_{x\to1}\Big(\dfrac{x^{4}+x^{2}-6}{x^{4}+2x+3}\Big)^{2}$
Apply the Limit of a Power law. The limit of a power is the power of the limit.
$\lim_{x\to1}\Big(\dfrac{x^{4}+x^{2}-6}{x^{4}+2x+3}\Big)^{2}=\Big(\lim_{x\to1}\dfrac{x^{4}+x^{2}-6}{x^{4}+2x+3}\Big)^{2}=...$
Since the limit of the denominator is not equal to $0$, the Limit of a Quotient law can be applied.
$...=\Big(\dfrac{\lim_{x\to1}x^{4}+x^{2}-6}{\lim_{x\to1}x^{4}+2x+3}\Big)^{2}=...$
Apply the Limit of a Sum and a Difference law in the numerator and the denominator.
$...=\Big(\dfrac{\lim_{x\to1}x^{4}+\lim_{x\to1}x^{2}-\lim_{x\to1}6}{\lim_{x\to1}x^{4}+\lim_{x\to1}2x+\lim_{x\to1}3}\Big)^{2}=...$
Apply the Limit of a Power law in the numerator and in the denominator again:
$...=\Big[\dfrac{(\lim_{x\to1}x)^{4}+(\lim_{x\to1}x)^{2}-\lim_{x\to1}6}{(\lim_{x\to1}x)^{4}+\lim_{x\to1}2x+\lim_{x\to1}3}\Big]^{2}=...$
Apply the Limit of a Constant Multiple law to the second term of the denominator:
$...=\Big[\dfrac{(\lim_{x\to1}x)^{4}+(\lim_{x\to1}x)^{2}-\lim_{x\to1}6}{(\lim_{x\to1}x)^{4}+2\lim_{x\to1}x+\lim_{x\to1}3}\Big]^{2}=...$
Evaluate:
$...=\Big[\dfrac{1^{4}+1^{2}-6}{1^{4}+2(1)+3}\Big]^{2}=\Big(\dfrac{1+1-6}{1+2+3}\Big)^{2}=\Big(\dfrac{-4}{6}\Big)^{2}=...$
$...=\Big(-\dfrac{2}{3}\Big)^{2}=\dfrac{4}{9}$
