How to solve this limit: $\lim_{x\to\infty}\sqrt{x+\sqrt{x}}-\sqrt{x-1}$ Compute the limit $\lim\limits_{x\to+\infty}\sqrt{x+\sqrt{x}}-\sqrt{x-1}$
my attempt:
I tried to multiply top and bottom by the conjugate
$$\begin{align}
\lim_{x\to+\infty}\sqrt{x+\sqrt{x}}-\sqrt{x-1}&=\lim_{x\to+\infty}\left(\sqrt{x+\sqrt{x}}-\sqrt{x-1}\right)\frac{\sqrt{x+\sqrt{x}}+\sqrt{x-1}}{\sqrt{x+\sqrt{x}}+\sqrt{x-1}}\\
&=\lim_{x\to+\infty}\frac{\left(\sqrt{x+\sqrt{x}}\right)^2-\left(\sqrt{x-1}\right)^2}{\sqrt{x+\sqrt{x}}+\sqrt{x-1}}\\
&=\lim_{x\to+\infty}\frac{(x+\sqrt{x})-(x-1)}{\sqrt{x+\sqrt{x}}+\sqrt{x-1}}\\
&=\lim_{x\to+\infty}\frac{1+\sqrt{x}}{\sqrt{x+\sqrt{x}}+\sqrt{x-1}}
\end{align}$$
But I don't know what I can do after this.
 A: Let's start from your last line:
$$\begin{align}
\lim_{x\to+\infty}\frac{1+\sqrt{x}}{\sqrt{x+\sqrt{x}}+\sqrt{x-1}} &= \lim \frac{\sqrt x}{\sqrt x} \frac{\frac{1}{\sqrt x} + 1}{\sqrt{1 + \frac{1}{\sqrt x}} + \sqrt{1 - \frac{1}{ x}}} \\
&= \frac{1}{1 + 1} = \frac{1}{2}
\end{align}$$
where we note that everywhere we have $\frac{1}{\sqrt x}$, those terms go to $0$ as $x \to \infty$. The method of factoring out the largest element in the numerator and denominator very often works. $\diamondsuit$
A: You may just observe that, as $x\to \infty$:
$$
\frac{1+\sqrt{x}}{\sqrt{x+\sqrt{x}}+\sqrt{x-1}}\sim \frac{\sqrt{x}\left(1+\frac1{\sqrt{x}}\right)}{\sqrt{x}\left(\sqrt{1+\frac{1}{\sqrt{x}}}\right)+\sqrt{x}\left(\sqrt{1-\frac{1}{x}}\right)}\sim \frac12.
$$
A: $$\begin{align}(x + x^{1/2})^{1/2} &= x^{1/2} + \frac 12 x^{-1/2}x^{1/2} -\frac 1 8 x^{-3/2}x+\cdots\\&=\sqrt x+\frac 12-\frac1{8\sqrt x} +\cdots\\
(x-1)^{1/2} &=\sqrt x -\frac 1{2\sqrt x}+\cdots \end{align}$$
therefore $$(x + x^{1/2})^{1/2} - (x-1)^{1/2}=\frac 1 2 + \frac 3{8\sqrt x} + \cdots \rightarrow \frac 12  \text{ as } x \to \infty.$$
A: We have $$\frac{1}{2}=\lim_{x\rightarrow\infty}\frac{\sqrt{x}}{2\sqrt{x}\left(\sqrt{\left(1+\frac{1}{\sqrt{x}}\right)}\right)}\leq\lim_{x\rightarrow\infty}\frac{\sqrt{x}}{\sqrt{x\left(1+\frac{1}{\sqrt{x}}\right)}+\sqrt{x}}\leq\lim_{x\rightarrow\infty}\frac{1+\sqrt{x}}{\sqrt{x+\sqrt{x}}+\sqrt{x-1}}\leq\lim_{x\rightarrow\infty}\frac{1+\sqrt{x}}{2\sqrt{x}}=\frac{1}{2}$$
A: Useful inequality chain
Given $a \ge 0$ and $b > -\frac{1}{2}a$:
  $a+b-2\frac{b^2}{a} \le a+b-\frac{b^2}{a+b} = \frac{a^2+2ab}{a+b} \le \sqrt{a^2+2ab} \le a+b-\frac{1}{2}\frac{b^2}{a+b} \le a+b$
Solution
Given $x \ge 1$:
  $\sqrt{x}+\frac{1}{2}-\frac{1}{4\sqrt{x}+2} \le \sqrt{x+\sqrt{x}} \le \sqrt{x}+\frac{1}{2}$.
  $\sqrt{x}-\frac{1}{2\sqrt{x}}-\frac{1}{4x\sqrt{x}} \le \sqrt{x-1} \le \sqrt{x}-\frac{1}{2\sqrt{x}}$.
Therefore $\sqrt{x+\sqrt{x}}-\sqrt{x-1} \to \frac{1}{2}$ as $x \to \infty$.
Notes
This kind of inequalities are useful when we do not want to use asymptotic expansion but otherwise as shown by abel expansion is the most widely applicable.
