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The given limit is:

\[
\lim_{{x \to 1}} \frac{x^5 + 3x^4 - 4}{\sqrt[5]{x} - 1}
\]

To solve this limit, we first substitute \(x = 1\) to check if the expression is indeterminate:

\[
\text{Numerator: } 1^5 + 3 \cdot 1^4 - 4 = 1 + 3 - 4 = 0
\]
\[
\text{Denominator: } \sqrt[5]{1} - 1 = 1 - 1 = 0
\]

Since both the numerator and denominator evaluate to zero, this is an indeterminate form of type \(\frac{0}{0}\). We can apply L'Hôpital's Rule or algebraic manipulation. Given the complexity of the denominator, L'Hôpital's Rule may be more straightforward.

### Step-by-step solution using L'Hôpital's Rule:

1. **Differentiate the numerator:**

   The numerator is \(x^5 + 3x^4 - 4\). Its derivative is:

   \[
   5x^4 + 12x^3
   \]

2. **Differentiate the denominator:**

   The denominator is \(\sqrt[5]{x} - 1 = x^{1/5} - 1\). Its derivative is:

   \[
   \frac{1}{5} x^{-4/5} = \frac{1}{5\sqrt[5]{x^4}}
   \]

3. **Apply L'Hôpital's Rule:**

   The limit now becomes:

   \[
   \lim_{{x \to 1}} \frac{5x^4 + 12x^3}{\frac{1}{5\sqrt[5]{x^4}}}
   \]

   Simplifying the denominator:

   \[
   \lim_{{x \to 1}} \left(5 \cdot (5x^4 + 12x^3) \cdot \sqrt[5]{x^4}\right) = \lim_{{x \to 1}} (25x^4 + 60x^3) \sqrt[5]{x^4}
   \]

4. **Evaluate the limit as \(x \to 1\):**

   When \(x = 1\):

   \[
   25 \cdot 1^4 + 60 \cdot 1^3 = 25 + 60 = 85
   \]
   
   And \(\sqrt[5]{1^4} = 1\).

5. **Conclusion:**

   \[
   \lim_{{x \to 1}} \frac{x^5 + 3x^4 - 4}{\sqrt[5]{x} - 1} = 85
   \]

### Final Answer:

\[
\boxed{85}
\]

Would you like more details on any step, or do you have any questions?

### Related Questions:

1. How does L'Hôpital's Rule apply to limits with indeterminate forms of type \(\frac{0}{0}\)?
2. What are alternative methods for solving limits when L'Hôpital's Rule is not applicable?
3. How can we simplify the expressions involving roots and exponents to find derivatives?
4. What are some examples where the direct substitution method works without L'Hôpital's Rule?
5. What other indeterminate forms exist, and how do we approach solving them?

### Tip:

Always check the form of a limit before applying any rule. If both the numerator and denominator tend to zero or infinity, consider L'Hôpital's Rule or factorization for simplification.

lim (x → 1) (x^5 + 3x^4 - 4) / (5th root(x) - 1)

Learn how to calculate the limit of (x^5 + 3x^4 - 4) / (5th root(x) - 1) as x approaches 1 using L'Hopital's Rule. This problem covers key calculus concepts like limits, indeterminate forms, and derivatives. Suitable for college-level students studying calculus.

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