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Solving: $\lim_{n\to\infty }\left(\sin\left(\pi \sqrt{4n^2+n}+2\cdot -\pi n\right)\right)$
Solve instead: $\lim_{x\to\infty}\sin\left(\pi\sqrt{4n^2+n}-2n\pi\right)$

Exercise

$\lim_{x\to\infty}\sin\left(\pi\sqrt{4n^2+n}-2n\pi\right)$

Step-by-step Solution

1

The limit $\lim_{n\to\infty }\left(\sin\left(\pi \sqrt{4n^2+n}+2\cdot -\pi n\right)\right)$ is indeterminate, because the limit of trigonometric functions as $n$ approaches infinity is indeterminate

indeterminate

Final answer to the exercise

indeterminate

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