3
$\begingroup$

Why is $$\lim_{x\to \infty} x\left( (x^2+1)^{\frac{1}{2}} - x\right) = \frac{1}{2}?$$

What is the right way to simplify this? My only idea is: $$x((x²+1)^{\frac{1}{2}} - x) > x(x^{2^{0.5}}) - x^2 = 0$$ But 0 is too imprecise.

$\endgroup$
4
  • 1
    $\begingroup$ others have shown you how to solve the problem. I can just add that you should be in general very careful with $\sqrt{x^2 +1}$ expression when 'pulling out' $x$: sign can be a problem, so make sure you write it as $|x| \sqrt{1+\frac{1}{x^2}}$ $\endgroup$
    – Alex
    Dec 25, 2014 at 15:44
  • $\begingroup$ Since x tends to infinity, $ (x²+1)^{\frac{1}{2}} - x $ should tend to zero as the product is finite. Rationalization is often helpful when you see such square root combination. $\endgroup$
    – Narasimham
    Dec 25, 2014 at 15:54
  • 1
    $\begingroup$ It's not correct to say that "$x$ becomes infinite". At every point while evaluating the limit, $x$ is finite. There is no point where $x$ is infinite. Instead, $x$ "approaches infinity", or "tends to infinity" might be more agreeable. While it's perfectly clear what you mean here, it's still important since language dictates thoughts. And you shouldn't think that $x$ is now $\infty$ when evaluating the limits. That's just not true. $\endgroup$
    – Asaf Karagila
    Dec 25, 2014 at 16:02

7 Answers 7

5
$\begingroup$

Rewrite the given term as $$x\frac{x^2+1-x^2}{\sqrt{x^2+1}+x}=\frac{x}{\sqrt{x^2+1}+x}$$ Assuming $x>0$, we have $$\frac{1}{\sqrt{1+\frac{1}{x^2}}+1}$$ This clearly tends to $\frac{1}{2}$ as $x\rightarrow\infty$.

$\endgroup$
4
$\begingroup$

You already know that $\sqrt{x^2 + 1} \sim x$. The problem is that that approximation isn't good enough for computing this limit.

However, you've done exercises to do this exact thing: use a differential approximation and the value of $\sqrt{x^2}$ to estimate the value of $\sqrt{x^2 + 1}$. e.g. maybe you've been asked to estimate $\sqrt{81.4}$ using the knowledge of $\sqrt{81}$.

To put it back in familiar symbols, set $f(t) = \sqrt{x^2 + t}$. Then differential approximation tells us that

$$ f(1) \approx f(0) + f'(0) \cdot (1 - 0) $$

or that

$$ \sqrt{x^2 + 1} \approx x + \frac{1}{2 \sqrt{x^2}} = x + \frac{1}{2x} $$

and now we can see that the limit probably should be $1/2$. There is one last problem, though: we need to make sure there isn't too much error in the differential approximation! And unfortunately, the description of the error term for differential approximation isn't good enough for our purposes, since it tells us what happens as $t \to 0$, not as $x \to \infty$.

While there are some tricks for doing this, I will instead give a more systematic approach: use Taylor series with remainder. Here is one form:

$$ f(1) = f(0) + f'(0) \cdot (1-0) + \frac{1}{2} f''(a) \cdot (1-0)^2 $$

where $a$ is some unknown value in the interval $[0,1]$. So we have

$$ \sqrt{x^2 + 1} = \sqrt{x^2} + \frac{1}{2 \sqrt{x^2} } - \frac{1}{8 (x^2 + a)^{3/2}} = x + \frac{1}{2 x } - \frac{1}{8 (x^2 + a)^{3/2}} $$

where $a \in [0,1]$ (and varies with $x$). Plugging this into our original limit now gives us something we can rigorously show converges to $1/2$.

It may help to use the fact this is a strictly increasing function of $a$ to arrange the above as

$$ x + \frac{1}{2 x } - \frac{1}{8x^3} \leq \sqrt{x^2 + 1} \leq x + \frac{1}{2 x } - \frac{1}{8(x^2 + 1)^{3/2}} $$


Interestingly, even though we needed a first-order estimate to solve the problem, we could have used a zeroth-order Taylor series with remainder, or equivalently the mean value theorem:

$$ f(1) = f(0) + f'(a) \cdot (1-0) = x + \frac{1}{2 \sqrt{x^2 + a}} $$

so that

$$x + \frac{1}{2\sqrt{x^2+1}} \leq \sqrt{x^2 + 1} \leq x + \frac{1}{2 x}$$

$\endgroup$
2
$\begingroup$

Hint: $\sqrt{x^2 + 1} - x$ can be written as $\sqrt{x^2 + 1} - \sqrt{x^2}$. Generally, if you have an expression involving $\sqrt a - \sqrt b$ then it is often useful to "expand" the expression with $\sqrt a + \sqrt b$:

$$ x \, (\sqrt{x^2 + 1} - x) = \frac {x \,(\sqrt{x^2 + 1} - x)(\sqrt{x^2 + 1} + x)} {\sqrt{x^2 + 1} + x} $$

$\endgroup$
0
1
$\begingroup$

It may help in doing the substitution $t=1/x$, so the limit becomes $$ \lim_{t\to0^+}\frac{\sqrt{\dfrac{1}{t^2}+1}-\dfrac{1}{t}}{t}= \lim_{t\to0^+}\frac{\sqrt{1+t^2}-1}{t^2} $$ Now recall that $\sqrt{1+x}=1+\frac{1}{2}x+o(x)$, so you get $$ \lim_{t\to0^+}\frac{1+\frac{1}{2}t^2+o(t^2)-1}{t^2}= \lim_{t\to0^+}\left(\frac{1}{2}+\frac{o(t^2)}{t^2}\right)=\frac{1}{2} $$ Alternatively, use l'Hôpital's theorem: $$ \lim_{t\to0^+}\frac{\sqrt{1+t^2}-1}{t^2} \overset{\textrm{(H)}}{=} \lim_{t\to0^+}\frac{\dfrac{t}{\sqrt{1+t^2}}}{2t} $$

$\endgroup$
1
$\begingroup$

the trouble with your approach stems from $0 \times \infty$ being indeterminate. here is an explanation.

(a) what you are doing is approximating $\sqrt{x^2+1} = x + \cdots$

(b) $\sqrt{x^2 + 1} - x = 0 + \cdots$

(c) $x(\sqrt{x^2+1} - x) = x(0 + \cdots) = 0?$

you think the answer is 0. but, perhaps $x \times \cdots \neq 0?$

now, i will go back and elaborate on that $\cdots$ by using binomial thorem $$(BIG + small)^{1/2} = BIG^{1/2} + {1 \over 2}BIG^{-1/2}small + \cdots$$

$(x^2+1)^{1/2} = x + {1 \over 2x} + \cdots$ now, if you go thru the steps (a)-(c) again, you see that $x(\sqrt{x^2+1} - x) = x({1 \over 2x}+ \cdots) = 1/2$

the reason is that $\cdots = const . 1/x^2 + \cdots$ and $x \times \cdots = 0$ as $x \to \infty$

$\endgroup$
1
$\begingroup$

Let $x=\tan\theta$.

Then $L=\displaystyle\lim_{x\to \infty} x( (x^2+1)^{\frac{1}{2}} - x ) \\= \displaystyle\lim_{{\theta\to \frac{\pi}{2}}^{-}} \tan\theta( \sec\theta - \tan\theta )\\=\displaystyle\lim_{{\theta\to \frac{\pi}{2}}^{-}} \frac{\sin\theta-\sin^2\theta}{\cos^2\theta}\\ =\displaystyle\lim_{{\theta\to \frac{\pi}{2}}^{-}} \frac{\sin\theta-\sin^2\theta}{1-\sin^2\theta}\\ =\displaystyle\lim_{{\theta\to \frac{\pi}{2}}^{-}} \frac{(1-\sin\theta)\sin\theta}{(1-\sin\theta)(1+\sin\theta)}\\ =\displaystyle\lim_{{\theta\to \frac{\pi}{2}}^{-}}\frac{\sin\theta}{1+\sin\theta}\\ =\dfrac{1}{2}.$

$\endgroup$
0
$\begingroup$

Let $$a=\lim_{x\to\infty}x\left(\sqrt{x^2+1}-x\right)$$ Now rationalize it: $$a=\lim_{x\to\infty}\dfrac{x\left(x^2+1-x^2\right)}{\sqrt{x^2+1}+x}$$ Then simplify it: $$a=\lim_{x\to\infty}\dfrac{x}{\sqrt{x^2+1}+x}$$ $$a=\lim_{x\to0}\dfrac{\dfrac1x}{\sqrt{\left(\dfrac1x\right)^2+1}+\dfrac1x}$$ $$a=\lim_{x\to0}\dfrac{1}{\sqrt{x^2+1}+1}$$ $$a=\dfrac{1}{\sqrt{0^2+1}+1}=\dfrac{1}{1+1}=\dfrac12$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .