Evaluate $\lim_{x\rightarrow \infty} (\sqrt{x^2 +ax} - \sqrt{x^2 +bx})$

Question

Evaluate $$\lim_{x\rightarrow \infty} (\sqrt{x^2 +ax} - \sqrt{x^2 +bx})$$

I tried the following:

$$\lim_{x\rightarrow \infty} (\sqrt{x^2 +ax} - \sqrt{x^2 +bx}) \cdot \frac{\sqrt{x^2 +ax} + \sqrt{x^2 +bx}}{\sqrt{x^2 +ax} + \sqrt{x^2 +bx}}$$

But ended up with

$$\lim_{x\rightarrow \infty} \frac{ax-bx}{\sqrt{x^2 +ax} + \sqrt{x^2 +bx}}$$

Which I'm not sure what to do with.

Answer 1

$$\lim_{x\rightarrow \infty} (\sqrt{x^2 +ax} - \sqrt{x^2 +bx}) \cdot \frac{\sqrt{x^2 +ax} + \sqrt{x^2 +bx}}{\sqrt{x^2 +ax} + \sqrt{x^2 +bx}} =(a-b)\lim_{x\rightarrow \infty}\frac{x}{\sqrt{x^2 +ax} - \sqrt{x^2 +bx}}=(a-b)\lim_{x\rightarrow \infty}\frac{x}{\vert{x}\vert\left(\sqrt{1+\frac{a}{x}}+\sqrt{1+\frac{b}{x}}\right)}=\frac{(a-b)}{2}\lim_{x\rightarrow \infty}\frac{x}{\vert x\vert}=\frac{a-b}{2}.$$

Using that $\vert x\vert=\sqrt{x^{2}}$.

Answer 2

$$\lim_{x\rightarrow \infty} \frac{ax-bx}{\sqrt{x^2 +ax} + \sqrt{x^2 +bx}}= \lim_{x\rightarrow \infty} \frac{ax-bx}{\sqrt{(x+\frac{a}{2})^2 -\frac{a^2}{4}} + \sqrt{(x+\frac{b}{2})^2 -\frac{b^2}{4}}}= \lim_{x\rightarrow \infty} \frac{ax-bx}{|x+\frac{a}{2}| + |x+\frac{b}{2}|} $$ this is $$\lim_{x\rightarrow +\infty} \frac{(a-b)x}{2x+\frac{a+b}{2}}=\color{blue}{\frac{a-b}{2}}$$ as $x\rightarrow +\infty$, and $$\lim_{x\rightarrow -\infty} \frac{(a-b)x}{-2x-\frac{a+b}{2}}=\color{blue}{-\frac{a-b}{2}}$$ as $x\rightarrow -\infty$.

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Of course I have done the first from you have been stucked, so you can do the main limit in this way $$\lim_{x\rightarrow \infty} \sqrt{(x+\frac{a}{2})^2 -\frac{a^2}{4}} - \sqrt{(x+\frac{b}{2})^2 -\frac{b^2}{4}}=\lim_{x\rightarrow \infty}|x+\frac{a}{2}| - |x+\frac{b}{2}|$$ and obtain the same results.

Answer 3

I made the subsitution $y=\frac{1}{x}$. The limit as $y \rightarrow 0$ falls straight out as $\frac{a-b}{2}$. You'll need to use binomial expansion of $\sqrt{1+p} = 1 + \frac{p}{2}+...$

Answer 4

We may suppose $x>0$, so \begin{align}\sqrt{x^2+ax}+\sqrt{x^2+bx}&=x\biggl(\sqrt{1+\frac ax}+\sqrt{1+\frac bx} \,\biggr)=x\biggl(1+\frac a{2x}+1+\frac b{2x} +o\Bigl(\frac1x\Bigr) \biggr)\\ &=2x+\frac{a+b}2+o(1),\end{align} so that the denominator is equivalent at $\infty$ to $2x$, and $$\sqrt{x^2 +ax} - \sqrt{x^2 +bx}=\frac{ax-bx}{\sqrt{x^2 +ax} + \sqrt{x^2 +bx}}\sim_\infty\frac{(a-b)x}{2x}=\frac{a-b}{2}.$$

Answer 5

another method: asymptotics
For $x>0$, we have as $x \to \infty$: $$ \sqrt{x^2+ax} = x\sqrt{1+\frac{a}{x}} = x\left(1+\frac{a}{2x}+o(x^{-1})\right) = x+\frac{a}{2}+o(1) \\ \sqrt{x^2+bx}= x+\frac{b}{2}+o(1) \\ \sqrt{x^2+ax}-\sqrt{x^2+bx}= \frac{a-b}{2}+o(1) $$