As said in the question, I am trying to evaluate the limit of \begin{equation} \frac{\sqrt{n^2 + 1} - n}{\sqrt{n^2 + 1} + n} \end{equation} as $n$ goes to infinity ($n \to \infty$).
Like many other limit questions that involve radicals, I first multiplied the original expression with a fraction having both its numerator and denominator as $\sqrt{n^2 - 1} + n$: \begin{align} \frac{\sqrt{n^2 + 1} - n}{\sqrt{n^2 + 1} + n} &= \frac{\sqrt{n^2 + 1} - n}{\sqrt{n^2 + 1} + n} \times \frac{\sqrt{n^2 + 1} - n}{\sqrt{n^2 + 1} - n} \\ &= \frac{(\sqrt{n^2 + 1} - n)^2}{(\sqrt{n^2 + 1} + n)(\sqrt{n^2 + 1} - n)} \end{align} Then, it was just basic arithmetic from this point: \begin{align} \frac{(\sqrt{n^2 + 1} - n)^2}{(\sqrt{n^2 + 1} + n)(\sqrt{n^2 + 1} - n)} &= \frac{(\sqrt{n^2 + 1})^2 - 2n\sqrt{n^2 + 1} + n^2}{(\sqrt{n^2 + 1})^2 - n^2} \\ &= \frac{(n^2 + 1) - 2n\sqrt{n^2 + 1} + n^2}{(n^2 + 1) - n^2} \\ &=2n^2 - 2n\sqrt{n^2 + 1} + 1 \end{align} So, \begin{equation} \frac{\sqrt{n^2 + 1} - n}{\sqrt{n^2 + 1} + n} = 2n^2 - 2n\sqrt{n^2 + 1} + 1 \end{equation} *Have double checked for correctness with online math tools.
Here is where my problem lies:
Clearly, $2n^2$ goes to $\infty$ and $-2n\sqrt{n^2 + 1}$ goes to $-\infty$. Hence, more must be done to this expression before using direct substitution.
My issue is, I am not able to manipulate this expression anymore.
Any hints or help towards possible directions would be genuinely appreciated.
