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$$\lim_{x\to\infty}\frac{x - \sqrt{x^2+5x+2}}{x-\sqrt{x^2+0.5x + 1}}.$$

Can't find a means to resolve. The answer is $10$ by graphing.

Ross Millikan
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Bob
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3 Answers3

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We give two solutions. The first uses a standard trick that you have probably already seen. The second uses the definition of the derivative.

You are familiar with the identity $$(\sqrt{u}-\sqrt{v})(\sqrt{u}+\sqrt{v})=u-v, \qquad \qquad (\ast)$$ at least in its incarnation $(a-b)(a+b)=a^2-b^2$. We will use $(\ast)$ twice. First, we multiply the top and bottom of your expression by $$(x +\sqrt{x^2+5x+2})(x+\sqrt{x^2+0.5x + 1}).$$ We get $$\frac{x - \sqrt{x^2+5x+2}}{x-\sqrt{x^2+0.5x + 1}}=\frac{(x - \sqrt{x^2+5x+2})(x +\sqrt{x^2+5x+2})(x+\sqrt{x^2+0.5x + 1})}{(x-\sqrt{x^2+0.5x + 1})(x+\sqrt{x^2+0.5x + 1})(x +\sqrt{x^2+5x+2})}.$$ The product of the first two terms on top is $-(5x+2)$, and the product of the first two terms at the bottom is $-(0.5x+1)$. It follows that $$\frac{x - \sqrt{x^2+5x+2}}{x-\sqrt{x^2+0.5x + 1}}=\frac{(5x+2)(x+\sqrt{x^2+0.5x + 1})}{ (0.5x+1)(x +\sqrt{x^2+5x+2})}.\qquad\qquad(\ast\ast)$$

The behaviour of the right-hand side of $(\ast\ast)$ for large $x$ is almost obvious. To do the calculation formally, divide top and bottom by $x^2$, as follows: $$\frac{(5x+2)(x+\sqrt{x^2+0.5x + 1})}{ (0.5x+1)(x +\sqrt{x^2+5x+2})}= \frac{(5+\frac{2}{x})(1+\sqrt{1+ \frac{0.5}{x} + \frac{1}{x^2}})}{ (0.5+\frac{1}{x})(1 +\sqrt{1+\frac{5}{x}+\frac{2}{x^2}})}.\qquad\qquad(\ast\ast\ast)$$ If we look at the right-hand side of $(\ast\ast\ast)$, it should be clear that as $x \to\infty$, the top approaches $10$ and the bottom approaches $1$.

Another way: Divide top and bottom by $x$. We get $$\frac{1 - \sqrt{ 1+\frac{5}{x}+\frac{2}{x^2}}} {1-\sqrt{ 1+\frac{0.5}{x} + \frac{1}{x^2} }}.$$ Make the substitution $t=\frac{1}{x}$, and change the sign of top and bottom. We are now interested in the limit as $t$ approaches $0$ from the right of $$\frac{\sqrt{1+5t+2t^2}-1} {\sqrt{1+0.5t + t^2 }-1}.$$ Looks nice than before! We could now use a trick much like in the first solution. But we will use another idea. Multiply top and bottom by $t-0$. We get $$\frac{\sqrt{ 1+5t+2t^2}-1}{t-0}\frac{t-0}{\sqrt{ 1+0.5t + t^2 }-1}.$$ What is $$\lim_{t\to 0}\frac{\sqrt{ 1+5t+2t^2}-1}{t-0}?$$ If we look back on the definition of the derivative, we will recognize this limit as the derivative of $f(t)=\sqrt{1+5t+2t^2}$ at $t=0$. Calculate $f'(0)$. It is $5/2$. The limit of the second term is $1/g'(0)$, where $g(t)=1+0.5t +t^2$. Calculate. We get $1/(0.5/2)$. The product of the two limits is $10$.

Comment: The technique (in another answer) that uses the Taylor expansion is more sophisticated, but no harder. In the long run it is much more useful.

André Nicolas
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Divide through by $x, \lim \limits_{x\to\infty}\frac{x - \sqrt{x^2+5x+2}}{x-\sqrt{x^2+0.5x + 1}}=\lim \limits_{x\to\infty}\frac{1 - \sqrt{1+5/x+2/x^2}}{1-\sqrt{1+0.5/x + 1/x^2}}$ then use the expansion around $x=\infty \lim \limits_{x\to\infty}\frac{1 - \sqrt{1+5/x+2/x^2}}{1-\sqrt{1+0.5/x + 1/x^2}} \approx \frac{1-[1+5/(2x)]}{1-[1+1/(4x)]}=10$ where the $\approx$ means here that the first orders are the same.

Quixotic
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Ross Millikan
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Completing the squares:

$x^2+5x+2 = (x + 2.5)^2 + O(1)$

$x^2+0.5x+1 = (x + 0.25)^2 + O(1)$

So:

$\sqrt{x^2+5x+2} = x + 2.5 + o(1)$

$\sqrt{x^2+0.5x+1} = x + 0.25 + o(1)$

Now the answer is immediate.

Updated to address didier's concerns:

We show using elementary methods that $\sqrt{u^2+O(1)}=u+o(1)$, which is the same as saying that for any constant $C$, $\sqrt{u^2+C} \to u$ as $u \to \infty$.

Lemma: If $|\epsilon| < 1$, then $1 - |\epsilon| \le \sqrt{1 + \epsilon} \le 1 + |\epsilon|$.

Proof: If $\epsilon \ge 0$, then $$1 - \epsilon \le 1 \le \sqrt{1 + \epsilon} \le 1 + \epsilon$$ because $\sqrt a \le a$ if $a \ge 1$.

If $\epsilon < 0$, put $\theta = -\epsilon$. Then

$$1 - |\epsilon| = 1 - \theta < \sqrt{1 - \theta} = \sqrt{1 + \epsilon}$$

because $a < \sqrt a$ if $0 < a < 1$. And

$$\sqrt{1+\epsilon} = \sqrt{1-\theta} < 1 < 1 + |\epsilon|$$

Corollary: If $|\epsilon| < 1$, then $|\sqrt{1+\epsilon} - 1| \le |\epsilon|$.

Now we just put $\epsilon = C/u^2$ in the corollary to get (for $u > \sqrt{|C|}$):

$$|\sqrt{u^2 + C} - u| = u\Big|\sqrt{1 + \frac{C}{u^2}} - 1\Big| \le u\frac{|C|}{u^2} = \frac{|C|}{u}$$

which tends to $0$ as $u \to \infty$.

TonyK
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  • Which takes for granted that $\sqrt{u^2+O(1)}=u+o(1)$, presumably when $u\to+\infty$. Which is true but rather on the sophisticated side for somebody asking the question (while we are at it, why not replace $o(1)$ by $O(1/u)$, which is equally true...). – Did Nov 29 '11 at 08:51
  • @Didier: My answer is much less sophisticated than the other two posted so far! And why use $O(1/u)$ when $o(1)$ is enough? – TonyK Nov 29 '11 at 09:04
  • Enough but not proven in your answer. And since $o(1)$ is not what the obvious rigorous proof produces, one cannot help to wonder how you know that $o(1)$ is simply correct (not to mention how this kind of implicit step can actually help the OP). – Did Nov 29 '11 at 09:32
  • @didier: Now it's proven :-) – TonyK Nov 29 '11 at 11:58
  • Much better now. But [for any constant $C$, $\sqrt{u^2+C}\to u$ as $u\to\infty$] is absurd. – Did Nov 29 '11 at 13:43
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    @Didier: You're being very strict today! "$f(x) \to g(x)$ as $x \to \infty$" means the same as "$f(x) - g(x) \to 0$ as $x \to \infty$". Obviously. – TonyK Nov 29 '11 at 13:50
  • Wonderful answer, much shorter and trickier than other ones. I like the way you used. – Gigili Nov 29 '11 at 15:15
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    The thing is, when a question got more than one answer, users up-vote the one which is posted by top users. "the more reputation you've earned, the more reliable is your answer", that's it. – Gigili Nov 29 '11 at 15:22
  • Obviously. Are you serious? For example, you would write that $1/x\to\mathrm e^{-x}$ when $x\to+\infty$? Wow. – Did Nov 29 '11 at 16:26
  • @Didier: I don't expect I would. I don't expect I would write "$1/x - e^{-x} \to 0$ when $x \to +\infty$" either. But they both mean the same thing, obviously. – TonyK Nov 29 '11 at 16:48
  • This is becoming more absurd by the minute, so I will stop after this comment. (1) There is no reason one could not write [$(1/x)-\mathrm e^{-x}\to0$ when $x\to+\infty$]: this statement has a rigorous definition (which is that $\lim\limits_{x\to+\infty}u(x)=0$, where $u:x\mapsto (1/x)-\mathrm e^{-x}$ for every $x\gt0$) and it is true. (2) No one uses $f(x)\to g(x)$ as a shorthand for $f(x)-g(x)\to0$. In fact, no one writes $f(x)\to g(x)$ (except when $g(x)$ does not depend on $x$). (3) To pretend .../... – Did Nov 29 '11 at 17:11
  • .../... that ["$f(x)\to g(x)$ as $x\to\infty$" means the same as "$f(x)-g(x)\to0$ as $x\to\infty$". Obviously.] and, simultaneously, to have qualms at writing [$1/x\to \mathrm e^{-x}$ as $x\to\infty$] does not strike me as a particularly consistent position. – Did Nov 29 '11 at 17:11
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    @Didier: If I have invented this particular shorthand (which I doubt), then I am proud of my achievement! It is unambiguous and useful. – TonyK Nov 29 '11 at 17:25
  • @Didier: "to have qualms at writing [$1/x\to \mathrm e^{-x}$ as $x\to\infty$]": I have no qualms about it at all. It's just not something I would expect to want to write. It seems to me that you deliberately misunderstood me. – TonyK Nov 29 '11 at 18:56