Is it really impossible?

Calculus Level 5

For $n$ integer, what is the value of $\displaystyle \lim_{n \to \infty}\cos(\pi\sqrt{4n^{2}+5n+2})$ ?

\frac{1}{2}

1 0

-\frac{\sqrt{2}}{2}

\infty

Does not exist

1 solution

Green Elephant
Feb 12, 2018

The solution is: $\boxed {-\frac{\sqrt{2}}{2}}$ .

At first, this limit may seem impossible, given that $\cos(\infty)$ does not exist. However, that is not the case if we consider the period of the function cosine. We know that $\cos(\alpha+2n\pi)=\cos(\alpha)\cos(2n\pi)-\sin(\alpha)\sin(2n\pi)=\cos(\alpha)$ , and therefore, we rewrite our limit as:

$\displaystyle \large \lim_{n \to \infty}\cos(\pi\sqrt{4n^{2}+5n+2}-2n\pi+2n\pi)$ .

From there, we proceed by applying the cosine formula and we get $\displaystyle \lim_{n \to \infty}\cos(\pi\sqrt{4n^{2}+5n+2}-2n\pi) \rightarrow \displaystyle \lim_{n \to \infty}\cos(\pi(\sqrt{4n^{2}+5n+2}-2n))$ . We rationalize and we get $\displaystyle \lim_{n \to \infty}\cos(\pi(\frac{4n^{2}+5n+2-4n^{2}}{\sqrt{4n^{2}+5n+2}+2n}))$ , which yields $\large \cos(\frac{5\pi}{4})$ and gives us the final result.

This only holds if n is assumed to be an integer (the 2pi*n part). Since this is not true for limits, the overall limit does not exist. On the other hand, if we defined the function as a sequence depending on n, then the limit would exist since taking the infinite limit of a sequence assumes that each subscript is an integer.

Max Ranis - 3 years, 3 months ago

Which is why I stated that n is an integer. I posted this problem in the topic "Limits of Sequences and Series", because indeed, if the cosine above was a function, then the limit would not exist.

Green Elephant - 3 years, 3 months ago

I keep getting 5pi/2

Thomas Kirtley - 1 year ago

Is it really impossible?

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