Find the $\lim\limits_{x\rightarrow - \infty}\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x}$

Question

The task is to find $$\lim_{x\rightarrow - \infty}\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x}$$

What I've tried is dividing both the numerator and the denominator by $x$, but I just can't calculate it completely.

I know it should be something easy I just can't see.

Thanks in advance.

Answer 1

Let $y=-x\to \infty$ then

$$\lim_{x\rightarrow - \infty}\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x}=\lim_{y\rightarrow \infty}\frac{\sqrt{y^2+a^2}-y}{\sqrt{y^2+b^2}-y}$$

and

$$\frac{\sqrt{y^2+a^2}-y}{\sqrt{y^2+b^2}-y}\frac{\sqrt{y^2+a^2}+y}{\sqrt{y^2+a^2}+y}\frac{\sqrt{y^2+b^2}+y}{\sqrt{y^2+b^2}+y}=\frac{a^2}{b^2}\frac{\sqrt{y^2+b^2}+y}{\sqrt{y^2+a^2}+y}\\=\frac{a^2}{b^2}\frac{\sqrt{1+b^2/y^2}+1}{\sqrt{1+a^2/y^2}+1}\to \frac{a^2}{b^2}$$

Answer 2

Hint:

Do divide top and bottom by $x$ as you have tried, using that for negative values of $x$ (remember that $x\to -\infty$): $$x=-\sqrt{x^2}$$

Answer 3

We can rewrite the fraction as follows \begin{align*} \frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x}&=\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x}\times\frac{\sqrt{x^2+a^2}-x}{\sqrt{x^2+a^2}-x}\times \frac{\sqrt{x^2+b^2}-x}{\sqrt{x^2+b^2}-x}\\[4pt] &=\frac{x^2+a^2-x^2}{x^2+b^2-x^2}\times\frac{\sqrt{x^2+b^2}-x}{\sqrt{x^2+a^2}-x} \end{align*} So \begin{align*} \lim_{x\to-\infty}\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x}&=\frac{a^2}{b^2}\times\lim_{x\to-\infty}\frac{\sqrt{x^2+b^2}-x}{\sqrt{x^2+a^2}-x}\\[4pt] &=\frac{a^2}{b^2}\times\lim_{x\to-\infty}\frac{\sqrt{\frac{x^2}{x^2}+\frac{b^2}{x^2}}-\frac x{|x|}}{\sqrt{\frac{x^2}{x^2}+\frac{a^2}{x^2}}-\frac x{|x|}}\\[4pt] &=\frac{a^2}{b^2}\times\frac{\sqrt{1}-(-1)}{\sqrt{1}-(-1)}\\[4pt] &=\frac{a^2}{b^2} \end{align*}

Answer 4

From $$\lim_{x\rightarrow - \infty}\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x}$$ one can factor an $x$ from each term as follows: \begin{align} \frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x} &= \frac{x \left(1 + \sqrt{1 + \frac{a^{2}}{x^2}} \right)}{x \left( 1 + \sqrt{1 + \frac{b^2}{x^2}} \right)} = \frac{1 + \sqrt{1 + \frac{a^{2}}{x^2}} }{ 1 + \sqrt{1 + \frac{b^2}{x^2}} }. \end{align} From here the limit can be taken or one can expand the one more time. Using \begin{align} \sqrt{1 + t} = 1 + \frac{t}{2} - \frac{t^2}{8} + \mathcal{O}(t^3) \end{align} then \begin{align} \frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x} &= \frac{2 + \frac{a^2}{2 x^2} - \frac{a^4}{8 x^4} + \mathcal{O}\left(\frac{1}{x^6}\right)}{2 + \frac{b^2}{2 x^2} - \frac{b^4}{8 x^4} + \mathcal{O}\left(\frac{1}{x^6}\right)} = 1 + \frac{a^2 - b^2}{2 x^2} + \frac{2 b^4 - a^2 b^2 - a^2}{8 x^4} + \mathcal{O}\left(\frac{1}{x^6}\right). \end{align} Upon taking the limit the result becomes \begin{align} \lim_{x \to \pm \infty}\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x} = \lim_{x \to \pm \infty} 1 + \frac{a^2 - b^2}{2 x^2} + \frac{2 b^4 - a^2 b^2 - a^2}{8 x^4} + \mathcal{O}\left(\frac{1}{x^6}\right) = 1. \end{align}