How to get the limit of the sequence $(\sqrt{n^2+n}-\sqrt{n^2-1})$?

Question

My Attempt $$ \sqrt{n^2+n} - \sqrt{n^2-1} = \sqrt{n+1} \, \bigl(\sqrt{n}-\sqrt{n-1} \bigr) $$ Then I tried to apply the sandwich theorem in some way but failed.

Important Note

Please do not solve the problem. Give hints only. Thanks!

Further Progress

Below is what I have got following the hint.
$\frac {\sqrt {1+\frac {1}{n}}}{1+\sqrt {1-\frac {1}{n}}}$ which should converge to $\frac {1}{2}$, but the answer (without any working) says $\frac {1}{3}$.
Indeed, the answer should be $\frac {1}{2}$. I have got everything. Thank you, everyone!

Your factorisation is not correct. Please check the RHS of your expression. — Doug, Jan 12 '24 at 16:56
have you tried to multiply and divide by $\sqrt{n^2+n}+\sqrt{n^2-1}$? — Sine of the Time, Jan 12 '24 at 16:57
When you have the limit of the difference of two roots you have to do the inverse rationalization — Math Attack, Jan 12 '24 at 17:03
Plugging in $n = 10^6$ gives $0.5000003749737516$, which seems to indicate that the limit is 1/2. Where did you get 1/3? — Dan, Jan 12 '24 at 19:44

score 3 · Answer 1 · answered Jan 13 '24 at 07:34

The genral method to this kind of problem is called $\textbf{Rationalization}$, which means:

$$\sqrt{n^2+n} - \sqrt{n^2-1}$$

$$= (\sqrt{n^2+n} - \sqrt{n^2-1}) \frac{\sqrt{n^2+n} + \sqrt{n^2-1}}{\sqrt{n^2+n} + \sqrt{n^2-1}}$$

Then, obviously we can rearrange the divisor part as:

$$\frac{(n^2+n) - (n^2-1)}{\sqrt{n^2+n} + \sqrt{n^2-1}}$$

$$= \frac{n-1}{\sqrt{n^2+n} + \sqrt{n^2-1}}$$

$$= \frac{n-1}{n} \frac{1}{\sqrt{1+\frac{1}{n}}+\sqrt{1-\frac{1}{n^2}}}$$

Until this step, it is clear that the limit of the whole expression is:

$$1 \cdot \frac{1}{1+1} = \frac{1}{2}$$

Hope this can help you.

score 1 · Answer 2 · answered Jan 12 '24 at 20:24

As a first step, we note that if $k$ is fixed, we have $$ \sqrt{n + k} - \sqrt{n} = \frac{k}{\sqrt{n+k} + \sqrt{n}} \to 0 \quad \mbox{as} \quad n \to \infty. $$

Therefore, $$ \sqrt{n^2 + n} - \left(n+\frac12\right) = \sqrt{n^2 + n} - \sqrt{n^2+n+\frac14} \to 0 $$ and $$ \sqrt{n^2-1} - n = \sqrt{n^2-1} - \sqrt{n^2} \to 1 $$ as $n \to \infty$, which means that for the requested limit we obtain $$ \lim_{n \to \infty} \left(\sqrt{n^2 + n} - \sqrt{n^2-1}\right) = \lim_{n\to \infty} \left(\left(n + \frac12\right) - n \right) = \frac12. $$.

score 0 · Answer 3 · answered Jan 12 '24 at 19:49

Some hints for squeezing.

$\sqrt{x}-\sqrt{y}=\frac{x-y}{\sqrt{x}+\sqrt{y}}$

$x\ge y \implies \frac{x-y}{2\sqrt{x}}\le \frac{x-y}{\sqrt{x}+\sqrt{y}}\le \frac{x-y}{2\sqrt{y}}$

OR

$\sqrt{x}-\sqrt{x-1}=\sqrt{x}(1-\sqrt{1-\frac{1}{x}})\approx \sqrt{x}[1-(1-\frac{1}{2x})]=\frac{1}{2\sqrt{x}}$

How to get the limit of the sequence $(\sqrt{n^2+n}-\sqrt{n^2-1})$?

3 Answers3