To find [tex]\( \lim _{x \rightarrow 1} \frac{\sqrt[5]{x}-1}{x-1} \)[/tex], we can use several techniques in calculus. Let's go through the detailed steps:
### Step 1: Direct Substitution
First, let's try substituting [tex]\( x = 1 \)[/tex] directly into the expression [tex]\(\frac{\sqrt[5]{x}-1}{x-1}\)[/tex].
[tex]\[
\frac{\sqrt[5]{1}-1}{1-1} = \frac{1-1}{1-1} = \frac{0}{0}
\][/tex]
This gives us an indeterminate form [tex]\(\frac{0}{0}\)[/tex], which means we need to use another method to evaluate the limit.
### Step 2: L'Hôpital's Rule
L'Hôpital's Rule can be used to evaluate [tex]\(\frac{0}{0}\)[/tex] indeterminate forms. It states that if we have a limit of the form [tex]\(\frac{0}{0}\)[/tex] or [tex]\(\frac{\infty}{\infty}\)[/tex], we can take the derivative of the numerator and the derivative of the denominator:
[tex]\[
\lim _{x \rightarrow c} \frac{f(x)}{g(x)} = \lim _{x \rightarrow c} \frac{f'(x)}{g'(x)}, \text{ provided this limit exists.}
\][/tex]
Here, [tex]\( f(x) = \sqrt[5]{x} - 1 \)[/tex] and [tex]\( g(x) = x - 1 \)[/tex].
First, find the derivative of [tex]\( f(x) = \sqrt[5]{x} \)[/tex]:
[tex]\[
f'(x) = \frac{1}{5} x^{-\frac{4}{5}}
\][/tex]
Next, find the derivative of [tex]\( g(x) = x - 1 \)[/tex]:
[tex]\[
g'(x) = 1
\][/tex]
Now apply L'Hôpital's Rule:
[tex]\[
\lim _{x \rightarrow 1} \frac{\sqrt[5]{x}-1}{x-1} = \lim _{x \rightarrow 1} \frac{\frac{1}{5}x^{-\frac{4}{5}}}{1}
\][/tex]
Simplify the expression:
[tex]\[
\frac{1}{5} \lim _{x \rightarrow 1} x^{-\frac{4}{5}}
\][/tex]
Since [tex]\( x^{-\frac{4}{5}} \)[/tex] is continuous at [tex]\( x = 1 \)[/tex], we can substitute [tex]\( x = 1 \)[/tex]:
[tex]\[
\frac{1}{5} \cdot 1^{-\frac{4}{5}} = \frac{1}{5} \cdot 1 = \frac{1}{5}
\][/tex]
### Conclusion
Thus, the limit is:
[tex]\[
\lim _{x \rightarrow 1} \frac{\sqrt[5]{x}-1}{x-1} = \frac{1}{5}
\][/tex]
