3
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ HInt: Consider the cases $b<1$, $b=1$ and $b>1$. $\endgroup$ Jul 18, 2017 at 8:40
  • $\begingroup$ @MundronSchmidt, What does that imply? Could you please elaborate? $\endgroup$
    – pi-π
    Jul 18, 2017 at 8:41
  • $\begingroup$ @blue_eyed_....: jibournet have already elaborated it in the answer below. $\endgroup$ Jul 18, 2017 at 8:44

6 Answers 6

8
$\begingroup$

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 $b=1$:

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

$\endgroup$
7
  • $\begingroup$ @ibounet I think for $b<1$ you are wrong. Try $b=-1$. See please my answer. It's a right answer. $\endgroup$ Jul 18, 2017 at 8:59
  • $\begingroup$ 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 $-\frac{a\sqrt{x}}{2x}\to 0$. $\endgroup$
    – zwim
    Jul 18, 2017 at 9:03
  • $\begingroup$ @MichaelRozenberg : I edited my answer. $\endgroup$
    – pitchounet
    Jul 18, 2017 at 9:03
  • $\begingroup$ @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. $\endgroup$
    – pitchounet
    Jul 18, 2017 at 9:07
  • 1
    $\begingroup$ @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. $\endgroup$
    – pitchounet
    Jul 18, 2017 at 9:11
3
$\begingroup$

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

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

$\endgroup$
2
  • $\begingroup$ Could you please elaborate? $\endgroup$
    – pi-π
    Jul 18, 2017 at 8:41
  • $\begingroup$ @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. $\endgroup$ Jul 18, 2017 at 8:44
2
$\begingroup$

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 .

$\endgroup$
1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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 $0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.