Answer :

Alright, let’s evaluate the limit [tex]\(\lim_{n \to \infty} \left(n^3 - \sqrt{2 n^4 + n^2 + 1}\right)\)[/tex].

1. Express the Dominant Term:
First, note that as [tex]\( n \)[/tex] approaches infinity, the term [tex]\( 2 n^4 + n^2 + 1 \)[/tex] inside the square root is dominated by the [tex]\( 2 n^4 \)[/tex] term because [tex]\( n^4 \)[/tex] grows much faster than [tex]\( n^2 \)[/tex] and the constant 1.

2. Simplify Inside the Square Root:
We can factor out [tex]\( n^4 \)[/tex] inside the square root:
[tex]\[ \sqrt{2 n^4 + n^2 + 1} = \sqrt{n^4 (2 + \frac{1}{n^2} + \frac{1}{n^4})} \][/tex]

3. Extract [tex]\( n^4 \)[/tex] from the Square Root:
Extracting [tex]\( n^4 \)[/tex] from under the square root, we get:
[tex]\[ \sqrt{n^4 (2 + \frac{1}{n^2} + \frac{1}{n^4})} = n^2 \sqrt{2 + \frac{1}{n^2} + \frac{1}{n^4}} \][/tex]

4. Approach the Limit:
As [tex]\( n \)[/tex] approaches infinity, the terms [tex]\(\frac{1}{n^2}\)[/tex] and [tex]\(\frac{1}{n^4}\)[/tex] go to zero:
[tex]\[ \sqrt{2 + \frac{1}{n^2} + \frac{1}{n^4}} \to \sqrt{2} \][/tex]

5. Rewrite the Expression:
Substituting back, the expression [tex]\( n^3 - \sqrt{2 n^4 + n^2 + 1} \)[/tex] approaches:
[tex]\[ n^3 - n^2 \sqrt{2} \][/tex]

6. Evaluate the Limit:
Now, we need to consider [tex]\( n^3 - n^2 \sqrt{2} \)[/tex] as [tex]\( n \)[/tex] approaches infinity. The term [tex]\( n^3 \)[/tex] grows much faster than [tex]\( n^2 \sqrt{2} \)[/tex], therefore the limit of the expression as [tex]\( n \)[/tex] approaches infinity is dominated by [tex]\( n^3 \)[/tex].

7. Conclusion:
Since [tex]\( n^3 \)[/tex] grows without bound faster than [tex]\( n^2 \sqrt{2} \)[/tex], the overall limit is:
[tex]\[ \lim_{n \to \infty} \left(n^3 - n^2 \sqrt{2}\right) = \infty \][/tex]

So, the limit is:
[tex]\[ \boxed{\infty} \][/tex]