Key Concepts and Formulas

• Limits at Infinity: When evaluating limits as $x \rightarrow \infty$, we often divide the numerator and denominator by the highest power of $x$ to simplify the expression. Terms of the form $\frac{c}{x^n}$ (where $c$ is a constant and $n > 0$) approach 0 as $x \rightarrow \infty$.
• Binomial Expansion: The binomial theorem states that $(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$. For specific cases like $(a+b)^6$ and $(a-b)^6$, terms often cancel out when added or subtracted.
  • $(a+b)^6 + (a-b)^6 = 2 \left[ \binom{6}{0} a^6 b^0 + \binom{6}{2} a^4 b^2 + \binom{6}{4} a^2 b^4 + \binom{6}{6} a^0 b^6 \right]$
• Algebraic Manipulation: Simplifying expressions involving radicals and powers is crucial. We can factor out common terms from under the radical.

Step-by-Step Solution

Let the given limit be $L$. $L = \lim\limits_{x \rightarrow \infty} \frac{(\sqrt{3 x+1}+\sqrt{3 x-1})^6+(\sqrt{3 x+1}-\sqrt{3 x-1})^6}{\left(x+\sqrt{x^2-1}\right)^6+\left(x-\sqrt{x^2-1}\right)^6} x^3$

Step 1: Prepare terms for limit evaluation. We need to manipulate the terms in the numerator and denominator so that we can evaluate the limit as $x \rightarrow \infty$. We will factor out the highest power of $x$ from the expressions under the square roots and from the terms within the parentheses.

For the numerator: Consider the term $(\sqrt{3x+1}+\sqrt{3x-1})^6$. $\sqrt{3x+1} = \sqrt{x(3+\frac{1}{x})} = \sqrt{x} \sqrt{3+\frac{1}{x}}$ $\sqrt{3x-1} = \sqrt{x(3-\frac{1}{x})} = \sqrt{x} \sqrt{3-\frac{1}{x}}$ So, $(\sqrt{3x+1}+\sqrt{3x-1})^6 = (\sqrt{x}(\sqrt{3+\frac{1}{x}}+\sqrt{3-\frac{1}{x}}))^6 = x^3 \left(\sqrt{3+\frac{1}{x}}+\sqrt{3-\frac{1}{x}}\right)^6$ Similarly, $(\sqrt{3x+1}-\sqrt{3x-1})^6 = x^3 \left(\sqrt{3+\frac{1}{x}}-\sqrt{3-\frac{1}{x}}\right)^6$ The numerator becomes: $x^3 \left[ \left(\sqrt{3+\frac{1}{x}}+\sqrt{3-\frac{1}{x}}\right)^6 + \left(\sqrt{3+\frac{1}{x}}-\sqrt{3-\frac{1}{x}}\right)^6 \right]$

For the denominator: Consider the term $(x+\sqrt{x^2-1})^6$. $\sqrt{x^2-1} = \sqrt{x^2(1-\frac{1}{x^2})} = x \sqrt{1-\frac{1}{x^2}}$ So, $(x+\sqrt{x^2-1})^6 = (x+x\sqrt{1-\frac{1}{x^2}})^6 = x^6 \left(1+\sqrt{1-\frac{1}{x^2}}\right)^6$ Similarly, $(x-\sqrt{x^2-1})^6 = x^6 \left(1-\sqrt{1-\frac{1}{x^2}}\right)^6$ The denominator becomes: $x^6 \left[ \left(1+\sqrt{1-\frac{1}{x^2}}\right)^6 + \left(1-\sqrt{1-\frac{1}{x^2}}\right)^6 \right]$

Now, substitute these back into the limit expression: $L = \lim\limits_{x \rightarrow \infty} \frac{x^3 \left[ \left(\sqrt{3+\frac{1}{x}}+\sqrt{3-\frac{1}{x}}\right)^6 + \left(\sqrt{3+\frac{1}{x}}-\sqrt{3-\frac{1}{x}}\right)^6 \right]}{x^6 \left[ \left(1+\sqrt{1-\frac{1}{x^2}}\right)^6 + \left(1-\sqrt{1-\frac{1}{x^2}}\right)^6 \right]} x^3$

Step 2: Simplify the expression by cancelling powers of $x$. $L = \lim\limits_{x \rightarrow \infty} \frac{x^6 \left[ \left(\sqrt{3+\frac{1}{x}}+\sqrt{3-\frac{1}{x}}\right)^6 + \left(\sqrt{3+\frac{1}{x}}-\sqrt{3-\frac{1}{x}}\right)^6 \right]}{x^6 \left[ \left(1+\sqrt{1-\frac{1}{x^2}}\right)^6 + \left(1-\sqrt{1-\frac{1}{x^2}}\right)^6 \right]}$ $L = \lim\limits_{x \rightarrow \infty} \frac{\left(\sqrt{3+\frac{1}{x}}+\sqrt{3-\frac{1}{x}}\right)^6 + \left(\sqrt{3+\frac{1}{x}}-\sqrt{3-\frac{1}{x}}\right)^6}{\left(1+\sqrt{1-\frac{1}{x^2}}\right)^6 + \left(1-\sqrt{1-\frac{1}{x^2}}\right)^6}$

Step 3: Evaluate the limit as $x \rightarrow \infty$. As $x \rightarrow \infty$, we have $\frac{1}{x} \rightarrow 0$ and $\frac{1}{x^2} \rightarrow 0$. Let $a = \sqrt{3+\frac{1}{x}}$ and $b = \sqrt{3-\frac{1}{x}}$. As $x \rightarrow \infty$, $a \rightarrow \sqrt{3}$ and $b \rightarrow \sqrt{3}$. The numerator becomes: $(\sqrt{3}+\sqrt{3})^6 + (\sqrt{3}-\sqrt{3})^6 = (2\sqrt{3})^6 + (0)^6 = (2\sqrt{3})^6$

Let $c = 1$ and $d = \sqrt{1-\frac{1}{x^2}}$. As $x \rightarrow \infty$, $c=1$ and $d \rightarrow \sqrt{1-0} = 1$. The denominator becomes: $(1+1)^6 + (1-1)^6 = (2)^6 + (0)^6 = 2^6$

So, the limit is: $L = \frac{(2\sqrt{3})^6}{2^6}$

Step 4: Calculate the final value. $L = \frac{2^6 (\sqrt{3})^6}{2^6} = (\sqrt{3})^6$ $(\sqrt{3})^6 = (3^{1/2})^6 = 3^{(1/2) \times 6} = 3^3 = 27$

Common Mistakes & Tips

• Incorrectly factoring out $x$: Ensure that when factoring out $x$ from terms like $\sqrt{x^2-1}$, you get $x\sqrt{1-\frac{1}{x^2}}$ and not just $x$. Also, remember that $\sqrt{x^2} = |x|$, but since $x \to \infty$, $x$ is positive, so $\sqrt{x^2}=x$.
• Algebraic errors in binomial expansion: While direct expansion is possible, recognizing that $(a+b)^n + (a-b)^n$ simplifies to terms with even powers of $b$ can be helpful. In this case, for the second term in both numerator and denominator, the second part of the sum becomes zero.
• Forgetting the $x^3$ term: The $x^3$ term in the original expression needs to be accounted for. In Step 1, we see it cancels out with $x^3$ factors from the numerator and $x^6$ from the denominator, leaving the correct ratio of terms.

Summary

The problem involves evaluating a limit at infinity of a complex expression. The strategy is to simplify the expression by factoring out the highest powers of $x$ from the terms within the radicals and the main parentheses. This allows us to substitute $x \rightarrow \infty$ and observe that terms with $1/x$ or $1/x^2$ go to zero. We then apply the binomial expansion concept, noting that the terms involving the difference of square roots go to zero. Finally, we simplify the remaining terms to arrive at the numerical value of the limit.

The final answer is 27.

The final answer is $\boxed{27}$.