I kind of stumbled on these 3 limits. I tried to solve 2 of them, but I don't know if it's correct and for the third one I don't have any idea where to start from. Could help me, please?

1. \lim_{x \to +\infty} \( 2+ \frac{4^n + (-5)^n}{7^n + 1} \)^{2n^3-n^2} = \lim_{x \to +\infty} \( 2 + \frac{5^n}{7^n} \times\frac{\frac{4^n}{5^n} + (-1)^n}{1 + \frac{1}{7^n}} \) ^{2n^3-n^2} = \lim_{x \to +\infty} 2^{2n^3-n^2} = 2^\infty = \infty

2. \lim_{x \to +\infty} \( \frac{n^3 + 4n + 1}{2n^3 + 5}\)^{\frac{-2n^4 +1}{n^4 + 3n + 1}} = (2^{-1})^{-2} = 2^2 = 4

\lim_{x \to +\infty} \frac{n^3 + 4n +1}{2n^3 + 5} = \frac{1}{2}= 2^{-1}
\lim_{x \to +\infty} \frac{-2n^4 + 1}{n^4 + 3n +1} = -2

3. \lim_{x \to +\infty}(cos(-2013))^n = ?

For number 3 you know that the -1 < cos(2013) < +1. What do you get if you raise a number whose absolute value is less than 1 with a large exponent?

Oh, that's easy, it's zero. Thank you!