The limit as x approaches infinity

Question

$$\lim_{x\to\infty}x\left(1-\sqrt{1+\frac1{2x}}\right)$$

Can anyone explain how to get this?

Answer 1

Hint: Multiply by $$\frac{1+\sqrt{1+\frac{1}{2x}}}{1+\sqrt{1+\frac{1}{2x}}}.$$ The top then simplifies nicely.

Answer 2

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 $h\ne0$

Answer 3

$$\lim_{x\to \infty}x(1-\sqrt{1+\frac{1}{2x}})\frac{1+\sqrt{1+\frac{1}{2x}}}{1+\sqrt{1+\frac{1}{2x}}}=\lim_{x\to \infty}-\frac12 \frac{1}{1+\sqrt{1+\frac{1}{2x}}}=-\frac14$$

Answer 4

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.