We are given the following limit expression to solve:

[ \lim_{x \to \infty} \left( 21 \cdot x^{819} \cdot \left( \sqrt[40]{x^{21} + 1} - \sqrt[40]{x^{21} - 1} \right) \right) ]

Step-by-Step Approach:

1. Factor the Difference of Roots: Let's define $f(x) = \sqrt[40]{x^{21} + 1} - \sqrt[40]{x^{21} - 1}$. At large values of $x$, both $x^{21} + 1$ and $x^{21} - 1$ are approximately equal to $x^{21}$. Hence, the difference of roots can be approximated using binomial expansion.

2. Binomial Approximation for Large $x$: Using the binomial expansion for large powers: $\sqrt[40]{x^{21} + 1} \approx x^{21/40} \left( 1 + \frac{1}{40 x^{21}} \right),$ and $\sqrt[40]{x^{21} - 1} \approx x^{21/40} \left( 1 - \frac{1}{40 x^{21}} \right).$

   Now, subtracting these approximations: $f(x) = \left( x^{21/40} \left( 1 + \frac{1}{40 x^{21}} \right) \right) - \left( x^{21/40} \left( 1 - \frac{1}{40 x^{21}} \right) \right).$ Simplifying: $f(x) = x^{21/40} \left( \frac{2}{40 x^{21}} \right) = \frac{2}{40} x^{21/40 - 21}.$ This simplifies further to: $f(x) = \frac{2}{40} x^{-819/40}.$

3. Substitute Back into the Original Expression: Now, substitute this approximation of $f(x)$ back into the original limit expression: $\lim_{x \to \infty} \left( 21 \cdot x^{819} \cdot \frac{2}{40} x^{-819/40} \right).$ Simplifying this: $= \lim_{x \to \infty} \left( 21 \cdot \frac{2}{40} \cdot x^{819 - 819/40} \right).$

4. Simplify the Exponent: The exponent simplifies as follows: $819 - \frac{819}{40} = 819 \left( 1 - \frac{1}{40} \right) = 819 \cdot \frac{39}{40} = 798.075.$

   Therefore, the expression becomes: $\lim_{x \to \infty} \left( 21 \cdot \frac{2}{40} \cdot x^{798.075} \right).$ Since the exponent is positive ($798.075 > 0$), as $x \to \infty$, the whole expression will tend to infinity.

Thus, the limit is:

$\boxed{\infty}$

Would you like more details on any specific steps, or further explanations?

Related Questions:

1. What happens if the exponent in the final expression is negative?
2. How would the answer change if a lower power of $x$ was in the equation?
3. What role does the binomial expansion play in simplifying limits?
4. Can the difference of large powers always be approximated by binomial expansion?
5. How would the limit change if the powers of $x$ in the original expression were swapped?

Tip: When dealing with limits involving large powers and roots, using binomial approximations helps simplify the expressions for easier computation.