Problem 1. Calculate:
a) $\lim _{x \rightarrow \infty}\left(\sqrt{\frac{1}{4} x^{2}+x+2}+\sqrt{\frac{1}{4} x^{2}+2 x+5}-\sqrt{x^{2}+x+1}\right)$;
b) $\lim _{n \rightarrow \infty} \sin \left(\pi \sqrt{n^{2}+3}\right)$.
Problem 1. Calculate:
a) $\lim _{x \rightarrow \infty}\left(\sqrt{\frac{1}{4} x^{2}+x+2}+\sqrt{\frac{1}{4} x^{2}+2 x+5}-\sqrt{x^{2}+x+1}\right)$;
b) $\lim _{n \rightarrow \infty} \sin \left(\pi \sqrt{n^{2}+3}\right)$.

## Grading Rubric.

a) The limit can also be written as:

(1p) $\lim _{x \rightarrow \infty}\left(\sqrt{\frac{1}{4} x^{2}+x+2}-\frac{1}{2} \sqrt{x^{2}+x+1}\right)+\lim _{x \rightarrow \infty}\left(\sqrt{\frac{1}{4} x^{2}+2 x+1}-\frac{1}{2} \sqrt{x^{2}+x+1}\right)$

$(2 p) \quad=\frac{3}{4}+\frac{7}{4}=\frac{5}{2}$.

(2p) b) Since, $\sin \left(\pi \sqrt{n^{2}+3}\right)=(-1)^{n} \cdot \sin \left(\pi \sqrt{n^{2}+3}-n \pi\right)$, we obtain:

(2p) $\lim _{n \rightarrow \infty} \sin \left(\pi \sqrt{n^{2}+3}\right)=(-1)^{n} \cdot \lim _{n \rightarrow \infty} \sin \left(\pi \sqrt{n^{2}+3}-n \pi\right)=0$.