A limit involving square roots Transcribed from photo
$$\require{cancel}
\lim_{n\to\infty}\sqrt{n+\tfrac12}\left(\sqrt{2n+1}-\sqrt{2n+3}\right)\\[12pt]
$$
$$
\lim_{n\to\infty}\sqrt{n+\tfrac12}\lim_{n\to\infty}\left(\sqrt{2n+1}-\sqrt{2n+3}\right)\frac{\sqrt{2n+1}+\sqrt{2n+3}}{\sqrt{2n+1}+\sqrt{2n+3}}\\[12pt]
$$
$$
\lim_{n\to\infty}\underbrace{\sqrt{\frac{\cancel{n}1}{\cancel{n}}+\cancelto{0}{\frac1{2n}}}}_1\lim_{n\to\infty}\frac{(2n+1)-(2n+3)}{\sqrt{2n+1}+\sqrt{2n+3}}\\[24pt]
$$
$$
\lim_{n\to\infty}\frac{(\cancel{2n}+1)-(\cancel{2n}+3)}{\sqrt{2n+1}+\sqrt{2n+3}}=\lim_{n\to\infty}\frac{-2}{\sqrt{2n+1}+\sqrt{2n+3}}\cdot\frac1{\sqrt{n}}\\[12pt]
$$
$$
\lim_{n\to\infty}\frac{\cancelto{0}{\frac2{\sqrt{n}}}}{\sqrt{2+\cancelto{0}{\frac1n}}+\sqrt{2+\cancelto{0}{\frac3n}}}=\frac0{2\sqrt2}=0\\[24pt]
$$
$$
0\cdot1=0
$$
I solved it, but i dont know if it is right.
I got $0$
 A: $$\lim _{x\to \infty }\left(\sqrt{x+\frac{1}{2}}\right)\left(\sqrt{2x+1}-\sqrt{2x+3}\right)$$
As you said, we can rewrite: 
$$\left(\sqrt{2x+1}-\sqrt{2x+3}\right)=\frac{-2}{\sqrt{2x+1}+\sqrt{2x+3}}$$
Try factoring the $x$
$$\frac{-2}{\sqrt{2x+1}+\sqrt{2x+3}}=\frac{-2}{\sqrt{\color{red}{x}(2+1/x)}+\sqrt{\color{red}{x}(2+3/x)}}=\frac{-2}{\color{red}{\sqrt{x}}\sqrt(2+1/x)+\color{red}{\sqrt{x}}\sqrt{2+3/x}}$$
Now factor the $\sqrt{x}$
$$\frac{-2}{\color{red}{\sqrt{x}}\sqrt(2+1/x)+\color{red}{\sqrt{x}}\sqrt{2+3/x}}=\frac{-2}{\color{red}{\sqrt{x}}(\sqrt{(2+1/x)}+\sqrt{2+3/x})}$$
Similarly, factor out the $x$ from $\sqrt{x+1/2}$.
This is pretty trivial, the answer is:
$$\sqrt{x}\sqrt{1+\frac{1}{2x}}$$
Great! Now let's put them together!
$$\lim_{x\to\infty} (\sqrt{x}\sqrt{1+\frac{1}{2x}})\cdot \frac{-2}{\sqrt{x}(\sqrt{(2+1/x)}+\sqrt{2+3/x})}$$
Oh well look at that, the $\sqrt{x}$ cancel! Now we are just left with:
$$\lim_{x\to\infty} (\sqrt{1+\frac{1}{2x}})\cdot \frac{-2}{(\sqrt{(2+1/x)}+\sqrt{2+3/x})}$$
Roughly speaking, since $1/\infty=0$, our limit simplies:
$$\lim_{x\to\infty} \sqrt{1}\cdot \frac{-2}{(\sqrt{2}+\sqrt{2})}=\frac{-2}{2\sqrt{2}}=\frac{-1}{\sqrt{2}}$$
A: $$\sqrt{n+\frac{1}{2}}\cdot (\sqrt{2n+1}-\sqrt{2n+3})
=\sqrt{n+\frac{1}{2}}\cdot\frac{-2}{\sqrt{2n+1}+\sqrt{2n+3}}\\
=\sqrt{n}\cdot \sqrt{1+\frac{1}{2n}}\cdot\frac{-2}{\sqrt{2n}\cdot (\sqrt{2n+1}+\sqrt{2n+3})}\\
=\sqrt{1+\frac{1}{2n}}\cdot\frac{-\sqrt{2}}{\sqrt{1+\frac{1}{2n}}+\sqrt{1+\frac{3}{2n}}}$$
Thus, the required limit is $1\cdot \frac{-\sqrt 2}{1+1}=\frac{-1}{\sqrt2}$
A: Here's how I solved . Have a look
A: Alternatively:
$$\lim_{n\to\infty}\sqrt{n+\tfrac12}\left(\sqrt{2n+1}-\sqrt{2n+3}\right)=$$
$$\frac{1}{\sqrt{2}}\lim_{n\to\infty}\sqrt{2n+1}\left(\sqrt{2n+1}-\sqrt{2n+3}\right)=$$
$$\frac{1}{\sqrt{2}}\lim_{n\to\infty}\left(2n+1-\sqrt{4n^2+8n+3}\right)=$$
$$\frac{1}{\sqrt{2}}\lim_{n\to\infty}\left(2n+1-2n\sqrt{1+\frac{2}{n}+\frac{3}{4n^2}}\right)=$$
$$\frac{1}{\sqrt{2}}\lim_{n\to\infty}\left(2n+1-2n\left(1+\frac{1}{n}+O(n^{-2}\right)\right)=$$
$$\frac{1}{\sqrt{2}}\lim_{n\to\infty}\left(-1+O(n^{-1})\right)=-\frac{1}{\sqrt{2}}.$$
