The limit as x approaches infinity $$\lim_{x\to\infty}x\left(1-\sqrt{1+\frac1{2x}}\right)$$
Can anyone explain how to get this?
 A: Hint: Multiply by
$$\frac{1+\sqrt{1+\frac{1}{2x}}}{1+\sqrt{1+\frac{1}{2x}}}.$$
The top then simplifies nicely. 
A: Setting  $\dfrac1{2x}=h,x=\dfrac1{2h}$
$$\lim_{x\to\infty}x\left(1-\sqrt{1+\frac1{2x}}\right)$$
$$=\lim_{h\to0}\frac{1-\sqrt{1+h}}{2h}=\lim_{h\to0}\frac{1-(1+h)}{2h(1+\sqrt{1+h})}$$
Now cancel $h$ as $h\ne0$
A: $$\lim_{x\to \infty}x(1-\sqrt{1+\frac{1}{2x}})\frac{1+\sqrt{1+\frac{1}{2x}}}{1+\sqrt{1+\frac{1}{2x}}}=\lim_{x\to \infty}-\frac12 \frac{1}{1+\sqrt{1+\frac{1}{2x}}}=-\frac14$$
A: Since you are looking for the behavior of $$x\left(1-\sqrt{1+\frac1{2x}}\right)$$ when $x$ goes to large values, you could remember that, when $y$ is small, Taylor expansion is $$\sqrt{1+y}= 1+\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right)$$ Now, replace $y$ by $\frac1{2x}$ and then $$x\left(1-\sqrt{1+\frac1{2x}}\right)\approx x\Big(1-(1+\frac1{4x}-\frac1{32x^2})\Big)=-\frac{1}{4}+\frac{1}{32 x}$$
Developing $$\sqrt{1+y}$$ to first order gives the limit; pushing to second order shows how the expression approaches the limit.
