To solve the limit
[tex]\[
\lim _{x \rightarrow 16} \frac{x^{1 / 4}-2}{x-16}
\][/tex]
we can approach it step-by-step.
1. Direct Substitution:
First, we try to directly substitute [tex]\( x = 16 \)[/tex] into the expression.
[tex]\[
\frac{16^{1 / 4}-2}{16-16}
\][/tex]
Since [tex]\( 16^{1/4} \)[/tex] is 2 (because [tex]\( 2^4 = 16 \)[/tex]), we get [tex]\( \frac{2 - 2}{0} = \frac{0}{0} \)[/tex], which is an indeterminate form. Thus, we need another method to evaluate this limit.
2. Factoring and Simplification:
We notice that evaluating the limit directly does not work, so we need to manipulate the expression to eliminate the indeterminate form.
Let's rewrite the numerator [tex]\( x^{1 / 4} - 2 \)[/tex].
Recognize that [tex]\( x^{1 / 4} - 2 \)[/tex] can be seen as akin to [tex]\(\sqrt[4]{x} - 2\)[/tex]. Consider the numerator as a difference of roots and note that we can relate it to a factor involving [tex]\( x - 16 \)[/tex].
Next, to resolve this indeterminate form, we factor or use a technique associated with limits like L'Hôpital's Rule, but a simple trick here is to consider [tex]\(\sqrt[4]{x}\)[/tex].
3. Apply L'Hôpital's Rule:
Since we have a [tex]\(\frac{0}{0}\)[/tex] indeterminate form, we can apply L'Hôpital's Rule. L'Hôpital's Rule states that for functions [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] that are differentiable and approach 0 as [tex]\( x \)[/tex] approaches a value [tex]\( a \)[/tex],
[tex]\[
\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)},
\][/tex]
if this limit exists.
Apply L'Hôpital's Rule:
Let [tex]\( f(x) = x^{1/4} - 2 \)[/tex] and [tex]\( g(x) = x - 16 \)[/tex].
Differentiate [tex]\( f(x) \)[/tex]:
[tex]\[
f'(x) = \frac{d}{dx}\left(x^{1/4}\right) = \frac{1}{4} x^{-3/4} = \frac{1}{4 \sqrt[4]{x^3}}
\][/tex]
Differentiate [tex]\( g(x) \)[/tex]:
[tex]\[
g'(x) = \frac{d}{dx}\left(x - 16\right) = 1
\][/tex]
Using L'Hôpital's Rule:
[tex]\[
\lim_{x \to 16} \frac{x^{1/4} - 2}{x - 16} = \lim_{x \to 16} \frac{f'(x)}{g'(x)} = \lim_{x \to 16} \frac{\frac{1}{4 \sqrt[4]{x^3}}}{1}
\][/tex]
Substitute [tex]\( x = 16 \)[/tex]:
[tex]\[
\frac{1}{4 \sqrt[4]{16^3}} = \frac{1}{4 \sqrt[4]{(2^4)^3}} = \frac{1}{4 \cdot 2^3} = \frac{1}{4 \cdot 8} = \frac{1}{32}
\][/tex]
Thus, the limit is:
[tex]\[
\lim_{x \rightarrow 16} \frac{x^{1 / 4}-2}{x-16} = \frac{1}{32}
\][/tex]
