# Evaluate the given limit: $\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})$

Evaluate the following limit. $$\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})$$

My Attempt: $$=\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})$$ $$=\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx}) \times \dfrac {\sqrt {x-a} + \sqrt {bx}}{\sqrt {x-a} + \sqrt {bx}}$$ $$=\lim_{x\to \infty} \dfrac {x-a-bx}{\sqrt {x-a} + \sqrt {bx}}.$$

How do I proceed?

• HInt: Consider the cases $b<1$, $b=1$ and $b>1$. Jul 18, 2017 at 8:40
• @MundronSchmidt, What does that imply? Could you please elaborate?
– pi-π
Jul 18, 2017 at 8:41
• @blue_eyed_....: jibournet have already elaborated it in the answer below. Jul 18, 2017 at 8:44

If $b \geq 0$ with $b \neq 1$:

$$\lim \limits_{x \to +\infty} \sqrt{x-a} - \sqrt{bx} = \lim \limits_{x \to +\infty} \sqrt{x}\Bigg( \sqrt{1 - \frac{a}{x}} - \sqrt{b} \Bigg) = \begin{cases} +\infty & \text{if } 0 \leq b < 1 \\ -\infty & \text{if } b > 1 \end{cases}$$

because, as $x \to +\infty$:

$$\Bigg( 1 - \frac{a}{x} \Bigg)^{1/2} - \sqrt{b} = 1 - \sqrt{b} - \frac{a}{2x} + o\Big( \frac{1}{x} \Big).$$

As a result:

$$\lim \limits_{x \to +\infty} \Bigg( 1 - \frac{a}{x} \Bigg)^{1/2} - \sqrt{b} = 1 - \sqrt{b}.$$

If $b=1$:

$$\lim \limits_{x \to +\infty} \sqrt{x-a} - \sqrt{bx} = 0.$$

• @ibounet I think for $b<1$ you are wrong. Try $b=-1$. See please my answer. It's a right answer. Jul 18, 2017 at 8:59
• Beware of quick conclusion in the case $b=1$, Same argument for $(1-\frac a{\sqrt{x}})^\frac 12-\sqrt{b}\to 1-\sqrt{b}$, yet limit is not $0$ when $b=1$. We still have to show that the term $-\frac{a\sqrt{x}}{2x}\to 0$.
– zwim
Jul 18, 2017 at 9:03
• @MichaelRozenberg : I edited my answer. Jul 18, 2017 at 9:03
• @zwim : In the case $b=1$, I do not see any quick conclusion. The continuity of the function $x \mapsto (1-x)^{1/2}$ at $x=0$ is enough to conclude. Jul 18, 2017 at 9:07
• @MichaelRozenberg : I do not see the point of considering $x \to -\infty$. My solution does not aim at addressing all the possible cases. I believe that the cases considered in my solution are correct. If your solution is indeed much better, I let the community be the judge of that. Jul 18, 2017 at 9:11

Now, if $b=1$ and $x\rightarrow+\infty$ then the limit is $0$.

If $b\neq1$ then the limit does not exist.

– pi-π
Jul 18, 2017 at 8:41
• @blue_eyed_ If $b=1$ then $\lim\limits_{x\rightarrow+\infty}\frac{x-a-x}{\sqrt{x-a}+\sqrt{x}}=\lim\limits_{x\rightarrow+\infty}\frac{-a}{\sqrt{x-a}+\sqrt{x}}=0$. The answer follows from your work. Jul 18, 2017 at 8:44

That is a good idea! Now, intuitively speaking, a square root grows slower than a linear function so you should get "some" infinity. However, it depends on the constants $a$ and $b$ what the result will look like.

E.g. if $a = 0$ and $b=2$, you would get \begin{align*} \lim_{x \rightarrow \infty} \frac{x - 2x}{\sqrt{x} + \sqrt{2x}} &= \lim_{x \rightarrow \infty} \frac{-x}{\sqrt{x} + \sqrt{2x}} \\ &= \lim_{x \rightarrow \infty} \frac{- x}{(1 + \sqrt{2}) \sqrt{x}} = - \lim_{x \rightarrow \infty} \frac{\sqrt{x}}{(1 + \sqrt{2}) } \\ &= - \infty. \end{align*} But for other choices of constants $a,b$ you might get $+\infty$. So your answer depends on their values.

In general, a way to proceed would be to use L'Hospital's rules to analyse limits of quotients .

An often good trick is to make the substitution $t=1/x$. However, in this case $t=1/\sqrt{x}$ seems better, because we get $$\lim_{t\to0^+}\frac{\sqrt{1-at^2}-\sqrt{b}}{t}$$ The numerator has limit $1-\sqrt{b}$, therefore we see that, if $0\le b<1$ the limit is $\infty$, whereas for $b>1$ the limit is $-\infty$.

If $b=1$, this is the derivative at $0$ of the function $f(t)=\sqrt{1-at^2}$. Since $$f'(t)=-\frac{at}{\sqrt{1-at^2}}$$ we have $f'(0)=0$.

In conclusion $$\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})= \begin{cases} \infty & 0\le b<1 \\[4px] 0 & b=1 \\[4px] -\infty & b>1 \end{cases}$$

Alternatively: $$\lim_\limits{x\to\infty} \sqrt{x-a}-\sqrt{bx}=\lim_\limits{x\to\infty} \sqrt{x}-\sqrt{bx}=(1-\sqrt{b})\lim_\limits{x\to\infty} \sqrt{x}=\begin{cases} 0, \ if \ b=1 \\ \infty, \ if \ b\ne 1. \end{cases}$$

Clearly, for $b<1$, $-\infty$ and for $b>1$, $\infty$ (because $\sqrt{1-a/x}-\sqrt b$ tends to the constant $1-\sqrt b$).

Then

$$\sqrt{x-a}-\sqrt{x}=\frac{x-a-x}{\sqrt{x-a}+\sqrt{x}}$$ and the limit is $0$.