Let's find the limit of the given expression as [tex]\( x \)[/tex] approaches infinity:
[tex]\[
\lim_{x \to \infty} \frac{\sqrt{x^6 + 8}}{4x^2 + \sqrt{3x^4 + 1}}
\][/tex]
### Step-by-Step Solution:
1. Examine the highest power of [tex]\( x \)[/tex] in the numerator and the denominator:
- In the numerator, the term inside the square root is [tex]\( x^6 + 8 \)[/tex]. As [tex]\( x \to \infty \)[/tex], [tex]\( x^6 \)[/tex] will dominate over 8, so the numerator behaves like [tex]\( \sqrt{x^6} = x^3 \)[/tex].
- In the denominator, we have two terms: [tex]\( 4x^2 \)[/tex] and [tex]\( \sqrt{3x^4 + 1} \)[/tex]. As [tex]\( x \to \infty \)[/tex], [tex]\( 3x^4 \)[/tex] will dominate over 1 under the square root, so the second term behaves like [tex]\( \sqrt{3x^4} = x^2\sqrt{3} \)[/tex].
2. Simplify the expression considering the dominant terms:
- The numerator simplifies to approximately [tex]\( x^3 \)[/tex] for large [tex]\( x \)[/tex].
- The denominator simplifies to approximately [tex]\( 4x^2 + x^2\sqrt{3} = x^2(4 + \sqrt{3}) \)[/tex].
3. Rewrite the limit using these simplified expressions:
[tex]\[
\lim_{x \to \infty} \frac{x^3}{x^2(4 + \sqrt{3})}
\][/tex]
4. Simplify the fraction:
[tex]\[
\frac{x^3}{x^2(4 + \sqrt{3})} = \frac{x^3}{x^2} \cdot \frac{1}{4 + \sqrt{3}} = \frac{x}{4 + \sqrt{3}}
\][/tex]
5. Evaluate the limit:
- As [tex]\( x \to \infty \)[/tex], [tex]\( x / (4 + \sqrt{3}) \)[/tex] will also approach infinity because the numerator [tex]\( x \)[/tex] grows without bound while [tex]\( (4 + \sqrt{3}) \)[/tex] is a constant.
Therefore, the limit of the given expression as [tex]\( x \)[/tex] approaches infinity is:
[tex]\[
\boxed{\infty}
\][/tex]
