\( \lim _{n \rightarrow \infty} \frac{\sqrt{a^{2}+n}-\sqrt{a^{2}-n}}{n} \)
Answer
\( \lim _{n \rightarrow \infty} \frac{2n}{n\left(\sqrt{a^{2}+n}+\sqrt{a^{2}-n}\right)} \) = \( \lim _{n \rightarrow \infty} \frac{2}{\sqrt{a^{2}+n}+\sqrt{a^{2}-n}} \) = \frac{2}{2a} \)
Step 1 ：\( \lim _{n \rightarrow \infty} \frac{\sqrt{a^{2}+n}-\sqrt{a^{2}-n}}{n} \) = \( \lim _{n \rightarrow \infty} \frac{\left(\sqrt{a^{2}+n}+\sqrt{a^{2}-n}\right)}{n}\left(\frac{\sqrt{a^{2}+n}-\sqrt{a^{2}-n}}{\left(\sqrt{a^{2}+n}+\sqrt{a^{2}-n}\right)}\right) \)
Step 2 ：\( \lim _{n \rightarrow \infty} \frac{\left(\sqrt{a^{2}+n}+\sqrt{a^{2}-n}\right)}{n}\left(\frac{\sqrt{a^{2}+n}-\sqrt{a^{2}-n}}{\left(\sqrt{a^{2}+n}+\sqrt{a^{2}-n}\right)}\right) \) = \( \lim _{n \rightarrow \infty} \frac{2n}{n\left(\sqrt{a^{2}+n}+\sqrt{a^{2}-n}\right)} \)
Step 3 ：\( \lim _{n \rightarrow \infty} \frac{2n}{n\left(\sqrt{a^{2}+n}+\sqrt{a^{2}-n}\right)} \) = \( \lim _{n \rightarrow \infty} \frac{2}{\sqrt{a^{2}+n}+\sqrt{a^{2}-n}} \) = \frac{2}{2a} \)
