Proof of limit of $\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+\frac{1}{2}}\right)=\frac{1}{2}$

Question

I just started Courant's "Differential and Integral Calculus" to really get a grip on the math I should need in engineering. I'm currently stuck on this limit, which is Exercise 7 in Chapter I - $5$ (Limit of a Sequence). I'm not used to doing limit proofs without the properties. I've tried to rewrite it in some different ways to no avail over the past few hours. At this point the book hasn't mentioned epsilon-N proofs so I'd like to stay away from that. If someone could give me some pointers or hints so I can move on I'd be thankful.

Answer 1

Hint: $$\sqrt{n+1}-\sqrt n = \frac{1}{\sqrt{n+1}+\sqrt n}.$$

Answer 2

Hint: Note that $$\sqrt{n+1}-\sqrt{n}=\left(\sqrt{n+1}-\sqrt{n}\right)\times\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{n}}$$

Answer 3

From:

$$(\sqrt{n+1}-\sqrt n) (\sqrt{n+\frac{1}{2}}) = \frac{\sqrt{n+\frac{1}{2}}}{\sqrt{n+1}+\sqrt n}$$ $$\frac{\sqrt{n+\frac{1}{2}}}{2\sqrt{n+1}} \lt \frac{\sqrt{n+\frac{1}{2}}}{\sqrt{n+1}+\sqrt n} \lt \frac{\sqrt{n+\frac{1}{2}}}{2\sqrt n}$$ for $n \gt0$. Also we have:

$$\frac{\sqrt{n+\frac{1}{2}}}{2\sqrt{n+1}}=1/2\sqrt{\dfrac{n+\frac{1}{2}}{n+1}}$$

and the right-hand side tends to $1/2$. Similarly for the other bound and so the limit is 1/2.