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lemon14 · Oct 13, 2013, 05:00 AM

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

ebaines · Oct 14, 2013, 05:52 AM

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?

lemon14 · Oct 14, 2013, 06:09 AM

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