# $\lim_{x \rightarrow \ + \infty}(\sqrt{x^2 + 2x} – \sqrt{x^2 – 7x})$

Find the limit: $$\lim_{x \rightarrow \ + \infty}(\sqrt{x^2 + 2x} – \sqrt{x^2 – 7x})$$

I did the following:

\begin{align}
(\sqrt{x^2 + 2x} – \sqrt{x^2 – 7x}) = \frac{(\sqrt{x^2 + 2x} – \sqrt{x^2 – 7x})}{1} \cdot \frac{(\sqrt{x^2 + 2x} + \sqrt{x^2 – 7x})}{(\sqrt{x^2 + 2x} + \sqrt{x^2 – 7x})}
\end{align}

I know the final answer is $\frac{9}{2}$. After multiplying by the conjugate, I see where the $9$ in the numerator comes from. I just can’t remember how I solved the rest of the problem.

#### Solutions Collecting From Web of "$\lim_{x \rightarrow \ + \infty}(\sqrt{x^2 + 2x} – \sqrt{x^2 – 7x})$"

$$\lim_{x\to \infty}\frac{(\sqrt{x^2+2x}-\sqrt{x^2-7x})(\sqrt{x^2+2x}+\sqrt{x^2-7x})}{\sqrt{x^2+2x}+\sqrt{x^2-7x}}$$
$$=\lim_{x\to \infty}\frac{9x}{\sqrt{x^2+2x}+\sqrt{x^2-7x}}=\lim_{x\to \infty}\frac{\frac{9x}{x}}{\frac{\sqrt{x^2+2x}}{x}+\frac{\sqrt{x^2-7x}}{x}}$$$$=\lim_{x\to \infty}\frac{9}{\sqrt{\frac{x^2+2x}{x^2}}+\sqrt{\frac{x^2-7x}{x^2}}}=\lim_{x\to \infty}\frac{9}{\sqrt{1+\frac 2x}+\sqrt{1-\frac 7x}}=\frac{9}{1+1}=\frac 92.$$

Expanding the rightmost term, we get:
$$\frac{(x^2+2x) – (x^2-7x)}{\sqrt{x^2+2x} + \sqrt{x^2-7x}} = \frac{9x}{\sqrt{x^2+2x} + \sqrt{x^2-7x}}$$
now just divide the numerator and denominator by $x = \sqrt{x^2}$:
$$\frac{9}{\sqrt{1+\frac{2}{x}} + \sqrt{1-\frac{7}{x}}}.$$
We have that $\sqrt{1+\frac{2}{x}} + \sqrt{1-\frac{7}{x}} \rightarrow 2$ as $x \rightarrow \infty$, so the final answer is $9/2$.

Setting $\dfrac1x=h,$ the limit reduces to $$\lim_{h\to0^+}\frac{\sqrt{1+2h}-\sqrt{1-7h}}h$$

$$=\lim_{h\to0^+}\frac{1+2h-(1-7h)}{h(\sqrt{1+2h}+\sqrt{1-7h})}$$

$$=\lim_{h\to0^+}\frac9{\sqrt{1+2h}+\sqrt{1-7h}}$$

$$=\frac9{\sqrt{1+2\cdot0}+\sqrt{1-7\cdot0}}$$

$\bf{My\; Solution::}$ Given $\displaystyle \lim_{x\rightarrow \infty}\left(\sqrt{x^2+2x}-\sqrt{x^2-7x}\right) = \lim_{x\rightarrow \infty}x\left\{\left(1+\frac{2}{x}\right)^{\frac{1}{2}}-\left(1-\frac{7}{x}\right)^{\frac{1}{2}}\right\}$

Now Using Binomial Expansion::

$\displaystyle \lim_{x\rightarrow \infty}\left\{\left(1+\frac{1}{2}\cdot \frac{2}{x}+\frac{1}{2}\cdot -\frac{1}{2}\cdot \frac{4}{2x^2}……+\infty\right)-\left(1-\frac{1}{2}\cdot \frac{7}{x}-\frac{1}{2}\cdot -\frac{1}{2}\cdot \frac{49}{2x^2}.+\infty\right)\right\}$

So $\displaystyle \lim_{x\rightarrow \infty}\left(\sqrt{x^2+2x}-\sqrt{x^2-7x}\right) = 1+\frac{7}{2} = \frac{9}{2}$

$$\lim_{x\rightarrow \infty} (\sqrt{x^2+2x} – \sqrt{x^2-7x})$$$$=\lim_{x\rightarrow \infty} \bigg(\sqrt{x^2+2x+1} – \sqrt{x^2-7x+\frac{49}{4}}\bigg)$$ $$=\lim_{x\rightarrow \infty} \bigg( x+1 – \bigg(x-\frac{7}{2}\bigg)\bigg) =\frac{9}{2}$$

The mean value theorem gives $f(x+h) = f(x) + f'(x)h + {1 \over 2} f”(\xi) h^2$, where $\xi \in [x,x+h]$ (adjusted appropriately for sign of $h$).

This gives $|\sqrt{1+\theta} – (1+ {1 \over 2} \theta) | \le {1 \over 8} \theta^2$.

We have (if $x>0$),$\sqrt{x^2+2x} – \sqrt{x^2-7x} = x(\sqrt{1+{2 \over x}}-\sqrt{1-{7 \over 7}})$.

Then $| \sqrt{1+{2 \over x}}-\sqrt{1-{7 \over x}} – {1 \over 2}{9 \over x}| \le {1 \over 8} (({2 \over x})^2+({7 \over x})^2)$.

Multiplying through by $x$ and letting $x \to \infty$ yields the desired result.

Note:
The point of this answer is that you can guess the answer by
using $\sqrt{1+\theta} \approx 1+ {1 \over 2} \theta$ and
writing
$\sqrt{1+{2 \over x}}-\sqrt{1-{7 \over x}} \approx {1 \over 2}{9 \over x}$.