Alright, let's evaluate the limit [tex]\(\lim _{x \rightarrow 16} \frac{x^{1 / 6}-2}{x^{1 / 2}-4}\)[/tex].
1. Understand the function:
The given function is [tex]\(\frac{x^{1/6} - 2}{x^{1/2} - 4}\)[/tex]. We need to determine what happens to this ratio as [tex]\(x\)[/tex] approaches 16.
2. Substitute [tex]\(x = 16\)[/tex]:
To find the limit, we start by substituting [tex]\(x = 16\)[/tex]:
[tex]\[ \frac{16^{1/6} - 2}{16^{1/2} - 4} \][/tex]
Simplify the expression:
- [tex]\(16^{1/2} = 4\)[/tex]
- [tex]\(16^{1/6} = 2\)[/tex] (since [tex]\((16^{1/2})^{1/3} = 4^{1/3}\)[/tex])
3. Evaluate the substitution:
Plugging these into the expression:
[tex]\[ \frac{2 - 2}{4 - 4} = \frac{0}{0} \][/tex]
Notice that this results in an indeterminate form ([tex]\(\frac{0}{0}\)[/tex]). Therefore, we need to apply further techniques to evaluate the limit.
4. Simplification approach (if possible):
We would typically look for ways to simplify the expression or apply L'Hôpital's Rule. However, analyzing the problem without resorting to complex transformations:
5. Behavior near the point of interest:
- As [tex]\(x \rightarrow 16\)[/tex], [tex]\(x^{1/2} - 4 \rightarrow 0\)[/tex] and [tex]\(x^{1/6} - 2 \rightarrow 0\)[/tex]
- Examine how fast [tex]\(x^{1/6}\)[/tex] and [tex]\(x^{1/2}\)[/tex] approach their respective limits.
6. L'Hôpital's Rule:
Given the [tex]\(\frac{0}{0}\)[/tex] form, we could use L'Hôpital's Rule, which suggests taking the derivative of the numerator and the denominator until the limit can be solved directly:
[tex]\[ \lim_{x \to 16} \frac{\frac{d}{dx}(x^{1/6} - 2)}{\frac{d}{dx}(x^{1/2} - 4)} \][/tex]
Calculate these derivatives:
[tex]\[ \frac{d}{dx}(x^{1/6} - 2) = \frac{1}{6} x^{-5/6} \][/tex]
[tex]\[ \frac{d}{dx}(x^{1/2} - 4) = \frac{1}{2} x^{-1/2} \][/tex]
Reevaluate the limit using these derivatives:
[tex]\[ \lim_{x \rightarrow 16} \frac{\frac{1}{6} x^{-5/6}}{\frac{1}{2} x^{-1/2}} = \lim_{x \rightarrow 16} \frac{\frac{1}{6}}{\frac{1}{2}} x^{(-5/6) - (-1/2)} \][/tex]
[tex]\[ = \frac{1}{3} \lim_{x \rightarrow 16} x^{(-5/6) + (1/2)} = \frac{1}{3} \lim_{x \rightarrow 16} x^{-1/3} \][/tex]
Simplify the power:
[tex]\[ x^{-1/3} = \frac{1}{x^{1/3}} \][/tex]
Substituting [tex]\(x = 16\)[/tex]:
[tex]\[ \frac{1}{3} \frac{1}{16^{1/3}} = \frac{1}{3} \cdot \frac{1}{2} \][/tex]
[tex]\[ = \frac{1}{6} \][/tex]
However, as x approaches 16, it becomes evident through a more rigorous evaluation (consider higher order terms and polynomial behavior) that the limit indeed tends to negative infinity due to the denominator approaches faster decline towards zero compared to the numerator.
So, the correct and final limit value for the given function, as x approaches 16, turns out to be:
[tex]\[ \lim _{x \rightarrow 16} \frac{x^{1 / 6}-2}{x^{1 / 2}-4} = -\infty \][/tex]
