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

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.

## closed as off-topic by user1729, Saad, user284331, JonMark Perry, user223391 Apr 7 '18 at 3:45

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user1729, Saad, user284331, JonMark Perry, Community
If this question can be reworded to fit the rules in the help center, please edit the question.

• Can you show us the result you get when you divided top and bottom by $x$? – Namaste Apr 4 '18 at 21:54
• when we deal with limit to $-\infty$ is often convenient to change the variable $x=-y$ in order to deal with a limit $\to +\infty$ – gimusi Apr 4 '18 at 22:01
• @gimusi you can also take advantage of the occasion to have your brain think in a way it is not used to - of course a mistake will happen more easily, but it is still a profit imo :) – Arnaud Mortier Apr 4 '18 at 22:07
• @ArnaudMortier I can only give my point of view on the best method to solve the limit and for my experience the change from $-\infty$ to $\infty$ is often advisable to avoid mistake. If you propose others method I'm happy to read about it and learn also different ways! – gimusi Apr 4 '18 at 22:11

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

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

• @ArnaudMortier When I face with limit to values different from $0$ and $\infty$ I always change the variable, not with the aim to simplify the expression but to in order avoid mistakes since handle limit to others values than 0 and $\infty$ often is more difficult. – gimusi Apr 4 '18 at 22:07
• I'm not arguing against that, it is definitely a very clever thing to do. But when you are used to thinking in some way, trying to do it in a different way does real good to the brain (physically). – Arnaud Mortier Apr 4 '18 at 22:13
• @ArnaudMortier In this case we can also use binomial expansion but the solution by radicals seems more simple to me in this case. – gimusi Apr 4 '18 at 22:16

Hint:

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

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*}

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}

• The mistake comes at the very first step. See my answer. – Arnaud Mortier Apr 9 '18 at 14:13