Limit to Infinity question?

Question

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

I'm not sure how to go about solving this problem.

Answer 1

Hints:

$\lim_{x\to \infty}(-\sqrt{-2x+x^2}+\sqrt{2x+x^2}) =\lim_{x\to \infty}\frac{(-\sqrt{-2x+x^2}+\sqrt{2x+x^2})\times (\sqrt{-2x+x^2}+\sqrt{2x+x^2})}{(\sqrt{-2x+x^2}+\sqrt{2x+x^2})}=?$

Then Do you know what to do next step?

In fact, $$=\lim_{x\to \infty}\frac{4x}{\sqrt{-2x+x^2}+\sqrt{2x+x^2}}= \lim_{x\to \infty}\frac{4}{\sqrt{-2\frac1x+1}+\sqrt{2\frac1x+1}}=2$$

Answer 2

$lim_{x \rightarrow \infty} \sqrt{2x + x^2} - \sqrt{-2x +x^2} =$ $lim_{x \rightarrow \infty} \sqrt{2x + x^2} - \sqrt{-2x +x^2} . \displaystyle\frac{\sqrt{2x + x^2} + \sqrt{-2x +x^2}}{\sqrt{2x + x^2} +\sqrt{-2x +x^2}} = $

$=lim_{x \rightarrow \infty} \displaystyle\frac{2x + x^2 - (-2x + x^2)}{\sqrt{2x + x^2} +\sqrt{-2x +x^2}}$

$= lim_{x \rightarrow \infty} \displaystyle\frac{4x}{ \sqrt{x^2(1 + 2/x)}+ \sqrt{x^2(1 - 2/x)}}$

$ = lim_{x \rightarrow \infty} \displaystyle\frac{4x}{ x \sqrt{(1 + 2/x)}+ x \sqrt{(1 - 2/x)}} $

$ = lim_{x \rightarrow \infty} \displaystyle\frac{4}{ \sqrt{(1 + 2/x)}+ \sqrt{(1 - 2/x)}} = 2 $

because

$= lim_{x \rightarrow \infty} \displaystyle\frac{1}{ \sqrt{(1 + 2/x)}+ \sqrt{(1 - 2/x)}} = 1/2$

Answer 3

Setting the limit to $0^+$ has often eased of my calculation

Setting $h=-\dfrac1x,$

$$F=\lim_{x\to \infty}(-\sqrt{-2x+x^2}+\sqrt{2x+x^2})$$

$$=\lim_{h\to0^+}\frac{-\sqrt{1-2h}+\sqrt{1+2h}}h$$

$$=\lim_{h\to0^+}\frac{1+2h-(1-2h)}{h(\sqrt{1-2h}+\sqrt{1+2h})}$$

Now cancelling $h$ as $h\ne0$ as $h\to0^+$

$$F=\lim_{h\to0^+}\frac4{\sqrt{1-2h}+\sqrt{1+2h}}$$

$$F=\lim_{h\to0^+}\frac4{\sqrt{1-2\cdot0}+\sqrt{1+2\cdot0}}=?$$

Answer 4

Another way using Taylor expansions : rewrite $$\sqrt{x^2+2x}-\sqrt{x^2-2x}=x\Big(\sqrt{1+\frac{2}{x}}-\sqrt{1+\frac{2}{x}}\Big)$$ and use the fact that, for small $y$ $$\sqrt{1+y}=1+\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right)$$ Replace $y$ by $\frac{2}{x}$ for the first radical and by $-\frac{2}{x}$ for the second radical.You so obtain $$\sqrt{x^2+2x}-\sqrt{x^2-2x}\approx x \times \frac{2}{x}=2$$ If you had used one extra term for the series expansion of $\sqrt{1+y}$, you would have arrived to $$\sqrt{x^2+2x}-\sqrt{x^2-2x}\approx2+\frac{1}{x^2}$$ which shows both the limit as well as the manner the limit is approached.

Answer 5

$\quad \lim_{x\to \infty}(-\sqrt{-2x+x^2}+\sqrt{2x+x^2})$

$=\lim_{x\to \infty}(-\sqrt{-2x+x^2+1}+\sqrt{2x+x^2+1})\\$ (because $\lim_{x\to\infty}\sqrt{N+1}-\sqrt{N}=0$) $=\lim_{x\to \infty}(-\sqrt{(x-1)^2}+\sqrt{(x+1)^2})$

$=\lim_{x\to \infty}(-{(x-1)}+{(x+1)})=2$