How do I calculate $\lim_{x\to+\infty}\sqrt{x+a}-\sqrt{x}$?

Question

I've seen a handful of exercises like this:

$$\lim_{x\to+\infty}(\sqrt{x+a}-\sqrt{x})$$

I've never worked with limits to infinity when there is some arbitrary number $a$. I am not given any details about it.

Apparently the answer is $0$. How was that conclusion reached?

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My guess is that since $x = +\infty$, the result of $x + a$ will still be $+\infty$ so we would have $\sqrt{x}-\sqrt{x} = 0$.

But that doesn't convince me. For starters, we don't know what $a$ is: it could be $-\infty$ or something, so $\infty - \infty$ would be indeterminate...

Answer 1

When an expression like that is written, the common assumption (in a Calculus course) is that $a$ is some real number.

The symbols $\infty$ of $-\infty$ do not represent real numbers and cannot be treated as such. Expressions like $x\to\infty$ and $\lim f(x)=\infty$ have precise meanings that do not involve the symbol $\infty$.

In this case, $$ \sqrt{x+a}-\sqrt x=(\sqrt{x+a}-\sqrt x)\,\frac{\sqrt{x+a}+\sqrt x}{\sqrt{x+a}+\sqrt x} =\frac{a}{\sqrt{x+a}+\sqrt x}. $$ For whatever number $a$ is, the expression on the right goes to zero as $x\to\infty$. To argue formally, note that $$ \left|\frac{a}{\sqrt{x+a}+\sqrt x}\right|=\frac{|a|}{\sqrt{x+a}+\sqrt x}\leq\frac{|a|}{\sqrt x}. $$ For a given $\varepsilon>0$ (that we think as small), if $x>a^2/\varepsilon^2$, then $$ \frac{|a|}{\sqrt x}<\frac{|a|}{|a|/\varepsilon}=\varepsilon. $$ This means that, if $x$ is big enough we can guarantee that $\frac{a}{\sqrt{x+a}+\sqrt x}$ it is as close to zero as we want.

Answer 2

You may write $$ \sqrt{x+a}-\sqrt{x}=\left(\sqrt{x+a}-\sqrt{x}\right)\frac{\sqrt{x+a} + \sqrt{x}}{\sqrt{x+a} + \sqrt{x}}=\frac{a}{\sqrt{x+a} + \sqrt{x}} $$ then it becomes easier to obtain your limit as $x \to \infty$.

Answer 3

Notice, $$\lim_{x\to \infty}(\sqrt{x+a}-\sqrt x)$$ $$=\lim_{x\to \infty}(\sqrt{x+a}-\sqrt x)\frac{(\sqrt{x+a}+\sqrt x)}{(\sqrt{x+a}+\sqrt x)}$$ $$=\lim_{x\to \infty}\frac{x+a-x}{\sqrt{x+a}+\sqrt x}$$ $$=a\lim_{x\to \infty}\frac{1}{\sqrt{x+a}+\sqrt x}=a(0)=0$$

Answer 4

METHOD 1:

We can write

$$\begin{align} \sqrt{x+a}&=\sqrt{x}\left(1+\frac ax\right)^{1/2}\\\\ &=\sqrt{x}\left(1+\frac{a}{2x}+O\left(\frac ax\right)^2\right)\\\\ &=\sqrt{x}+\frac{a}{2\sqrt{x}}+O\left(\frac 1x\right)^{3/2} \end{align}$$

Thus,

$$\sqrt{x+a}-\sqrt{x}=\frac{a}{2\sqrt{x}}+O\left(\frac 1x\right)^{3/2}\to 0$$

as $x\to \infty$.

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METHOD 2:

$$\begin{align} \sqrt{x+a}-\sqrt{x}&=\left(\sqrt{x+a}-\sqrt{x}\right)\,\frac{\sqrt{x+a}+\sqrt{x}}{\sqrt{x+a}+\sqrt{x}}\\\\ &=\frac{a}{\sqrt{x+a}+\sqrt{x}}\to 0 \end{align}$$

as $x\to \infty$.