The value of `lim_(xrarr1)(root5(x^(2))-2root5x+1)/(4(x-1)^(2))` is equal to

Answer

To solve the limit \( \lim_{x \to 1} \frac{\sqrt[5]{x^2} - 2\sqrt[5]{x} + 1}{4(x-1)^2} \), we will follow these steps:

### Step 1: Substitute \( x = 1 \)
First, we substitute \( x = 1 \) into the limit expression to check if we get an indeterminate form.

\[ \sqrt[5]{1^2} - 2\sqrt[5]{1} + 1 = 1 - 2 + 1 = 0 \]
\[ 4(1-1)^2 = 4 \cdot 0 = 0 \]

Since we get the form \( \frac{0}{0} \), we can apply L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule
According to L'Hôpital's Rule, we differentiate the numerator and the denominator separately.

**Numerator:**
Let \( f(x) = \sqrt[5]{x^2} - 2\sqrt[5]{x} + 1 \).

To differentiate \( f(x) \):
- The derivative of \( \sqrt[5]{x^2} \) is \( \frac{2}{5} x^{-\frac{3}{5}} \).
- The derivative of \( -2\sqrt[5]{x} \) is \( -\frac{2}{5} x^{-\frac{4}{5}} \).

Thus, the derivative of the numerator is:
\[ f'(x) = \frac{2}{5} x^{-\frac{3}{5}} - \frac{2}{5} x^{-\frac{4}{5}} \]

**Denominator:**
Let \( g(x) = 4(x-1)^2 \). The derivative is:
\[ g'(x) = 8(x-1) \]

### Step 3: Rewrite the limit
Now we can rewrite the limit using the derivatives:
\[ \lim_{x \to 1} \frac{f'(x)}{g'(x)} = \lim_{x \to 1} \frac{\frac{2}{5} x^{-\frac{3}{5}} - \frac{2}{5} x^{-\frac{4}{5}}}{8(x-1)} \]

### Step 4: Substitute \( x = 1 \) again
Now we substitute \( x = 1 \) in the new limit:
- For the numerator:
\[ f'(1) = \frac{2}{5}(1)^{-\frac{3}{5}} - \frac{2}{5}(1)^{-\frac{4}{5}} = \frac{2}{5} - \frac{2}{5} = 0 \]

- For the denominator:
\[ g'(1) = 8(1-1) = 0 \]

Again, we have the form \( \frac{0}{0} \). We need to apply L'Hôpital's Rule again.

### Step 5: Differentiate again
Differentiate \( f'(x) \) and \( g'(x) \) again.

**Numerator:**
\[ f''(x) = -\frac{6}{25} x^{-\frac{8}{5}} + \frac{8}{25} x^{-\frac{9}{5}} \]

**Denominator:**
\[ g''(x) = 8 \]

### Step 6: Rewrite the limit again
Now we can rewrite the limit:
\[ \lim_{x \to 1} \frac{f''(x)}{g''(x)} = \lim_{x \to 1} \frac{-\frac{6}{25} x^{-\frac{8}{5}} + \frac{8}{25} x^{-\frac{9}{5}}}{8} \]

### Step 7: Substitute \( x = 1 \) one last time
Now substitute \( x = 1 \):
\[ f''(1) = -\frac{6}{25}(1)^{-\frac{8}{5}} + \frac{8}{25}(1)^{-\frac{9}{5}} = -\frac{6}{25} + \frac{8}{25} = \frac{2}{25} \]
\[ g''(1) = 8 \]

### Step 8: Calculate the limit
Now we can calculate the limit:
\[ \lim_{x \to 1} \frac{\frac{2}{25}}{8} = \frac{2}{200} = \frac{1}{100} \]

Thus, the value of the limit is:
\[ \boxed{\frac{1}{100}} \]