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I am trying to focus on the limits of functions with similar series expansions and I stumbled on this. $$\lim_{x\to\infty}\left({\left(\frac{x^2+5}{x+5}\right)}^{1/2}\sin{\left({\left(\frac{x^3+5}{x+5}\right)}^{1/2}\right)}-(x-5)^{1/2}\sin{\left({\left(x^2-5x+25\right)}^{1/2}\right)}\right)=0$$

I heard the mean value is possible but the entire function is not bounded. I can take the taylor series at infinity however the terms would be undefined. I could use substitution with the taylor series but it would become a complicated mess.

$$\lim_{x\to\infty}\left(\left(\frac{x^2+5}{x+5}\right)^{1/2}{\left(\frac{x^3+5}{x+5}\right)}^{1/2}-\frac{\left(\frac{x^2+5}{x+5}\right)^{1/2}\left({\left(\frac{x^3+5}{x+5}\right)}^{1/2}\right)^3}{3!}+\frac{\left(\frac{x^2+5}{x+5}\right)^{1/2}\left({\left(\frac{x^3+5}{x+5}\right)}^{1/2}\right)^5}{5!}...-\left(\left({x-5}\right)^{1/2}{\left(x^2-5x+25\right)}^{1/2}-\frac{\left(x-5\right)^{1/2}\left({\left(x^2-5x+25\right)}^{1/2}\right)^3}{3!}+\frac{\left(x-5\right)^{1/2}\left({\left(x^2-5x+25\right)}^{1/2}\right)^5}{5!}....\right)\right)$$

I only have limited knowledge of series expansion so I am not so sure how to approach this. Is their an easier way?

Arbuja
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  • Somebody just woke up really, really wicked the day (s)he came up with this horror... – Timbuc Jul 13 '15 at 21:34
  • Nice exercise actually, forces you to think before embarking in random computations. First hint: how would you estimate the size of $$\sin{\left({\left(\frac{x^3+5}{x+5}\right)}^{1/2}\right)}-\sin{\left({\left(x^2-5x+25\right)}^{1/2}\right)}\ ?$$ This is $a(u(x)+v(x))-a(u(x))$ with $a(t)=\sin\sqrt{t}$, $u(x)=x^2-5x+25$ and $v(x)=$ $___$, hence... – Did Jul 13 '15 at 21:37
  • I was able to come up with a shortcut for this limit but I must prove it the HARD way. – Arbuja Jul 13 '15 at 21:37
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    ?? Not to underestimate the value of masochism... but how do you define "the hard way"? – Did Jul 13 '15 at 21:38
  • I have to prove this with series expansion or any method if possible? – Arbuja Jul 13 '15 at 21:39
  • I know the sine values converge according to the mean value theorem and is smaller compared to the polynomials that were multiplied but I'm just not sure how to truly solve this? – Arbuja Jul 13 '15 at 21:43
  • Second hint: The only required property of sine is to be a bounded function with a bounded derivative. – Did Jul 13 '15 at 21:47
  • I know to take the derivatives of sine and join the terms inside the brackets. I don't know how to deal with the multiplied polynomials. – Arbuja Jul 13 '15 at 21:52
  • Anyone willing to check? – Arbuja Jul 15 '15 at 00:43

3 Answers3

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Since $\sqrt{a}-\sqrt{b}=\frac{a-b}{\sqrt{a}+\sqrt{b}}$, we have $|\sqrt{a}-\sqrt{b}|\le\frac{|a-b|}{2\sqrt{\min(a,b)}}$. Therefore $$ \begin{align} \left|\left(\frac{x^3+5}{x+5}\right)^{1/2}-\left(\frac{x^3+125}{x+5}\right)^{1/2}\right| &\le\overbrace{\frac12\left(\frac{x^3+5}{x+5}\right)^{-1/2}}^{\large\frac1{2\sqrt{\min(a,b)}}}\overbrace{\frac{120}{x+5}\vphantom{\left(\frac{x^3}{x}\right)^{1/2}}}^{|a-b|}\\ &\le\frac{60}{x^2}\tag{1} \end{align} $$ Since $|\sin(x)-\sin(y)|\le|x-y|$, we get that $$ \left|\sin\left(\left(\frac{x^3+5}{x+5}\right)^{1/2}\right)-\sin\left(\left(\frac{x^3+125}{x+5}\right)^{1/2}\right)\right|\le\frac{60}{x^2}\tag{2} $$ Similarly, $$ \begin{align} \left|\left(\frac{x^2+5}{x+5}\right)^{1/2}-\left(\frac{x^2-25}{x+5}\right)^{1/2}\right| &\le\overbrace{\frac12(x-5)^{-1/2}}^{\large\frac1{2\sqrt{\min(a,b)}}}\overbrace{\frac{30}{x+5}}^{|a-b|}\\ &\le\frac{15}{(x-5)^{3/2}}\tag{3} \end{align} $$ Using $(2)$ and $(3)$ and $ab-cd=(a-c)b+(b-d)c$, we get $$ \begin{align} &\left|\left(\frac{x^2+5}{x+5}\right)^{1/2}\sin\left(\left(\frac{x^3+5}{x+5}\right)^{1/2}\right)-\left(\frac{x^2-25}{x+5}\right)^{1/2}\sin\left(\left(\frac{x^3+125}{x+5}\right)^{1/2}\right)\right|\\ &\le\left(\frac{x^2+5}{x+5}\right)^{1/2}\frac{60}{x^2}+1\cdot\frac{15}{(x-5)^{3/2}}\\[12pt] &\to0\tag{4} \end{align} $$

robjohn
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  • Rob John I appreciate your help and your approach appears correct but you haven't place the correct terms with the limits. If you look at my post on the top your final step has the limit different from mine. It seems rushed and I'm not sure how you got the bound in the first step. – Arbuja Jul 15 '15 at 00:35
  • I fear the second term at the end tends to $\infty$ – Piquito Jul 15 '15 at 02:18
  • @Ataulfo: that was a typo. The minus sign was for the numerator. In the denominator, there shouldn't have been a minus. Both terms are $O\left(\frac1{x^{3/2}}\right)$ – robjohn Jul 15 '15 at 02:53
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    @Arbuja: do you mean $(1)$? I have added a bit more explanation. – robjohn Jul 15 '15 at 03:28
  • @Arbuja: I am not sure what you mean: what terms are incorrect? – robjohn Jul 15 '15 at 06:26
  • The limit that I posted was $\lim_{x\to\infty}\left(\frac{x^2+5}{x+5}\right)\sin{\frac{x^3+5}{x+5}} -(x-5)\sin{\left(x^2-5x+25\right)}$. Thanks for the edits. – Arbuja Jul 15 '15 at 10:45
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    @Arbuja: Other than the square roots that are in the question and my answer, that is exactly what I have. Notice that $x-5=\frac{x^2-25}{x+5}$ and $x^2-5x+25=\frac{x^3+125}{x+5}$. – robjohn Jul 15 '15 at 11:54
  • @robjohn thank you for clarifying this to me. – Arbuja Jul 15 '15 at 12:10
  • @robjohn: Well for you. I had only read the typo when I made my comment. Regards. – Piquito Jul 15 '15 at 13:16
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We apply the following scheme $$AB-CD=\frac {\frac {1}{CD}-\frac {1}{AB}}{\frac{1}{ABCD}}$$ in which one has a shape which allows the application of the Hôpital´s rule. The numerator is $$N=\frac {1}{(\sqrt {x-5})sin\sqrt{x^2-5x+25}}-\frac {1}{\sqrt \frac {x^2+5}{x+5}sin\sqrt{\frac{x^3+5}{x+5}}}$$ and the denominator is $$D=\frac{1}{\sqrt\frac {(x-5)(x^2+5)}{(x+5)^2}sin\frac{x^3+5}{x+5}sin\sqrt{x^2-5x+25}}$$ The derivatives give (we use for short $\alpha$ and $\beta$ for the angles): $$N’=-\frac{1}{2(x-5)^{\frac {3}{2}}sin(\beta)} -\frac{(2x-5)}{2\sqrt{(x-5)(x^2-5x+25)}sin(\beta)}-\frac{1}{2\sqrt{(x+5)(x^2+5)}sin(\alpha)}+\frac{x\sqrt{x+5}}{(x^2+5)^{\frac {3}{2}}sin(\alpha)}+\frac{x^2+10x-5}{2(x+5)\sqrt{(x^2+5)(x^3+5)}}\frac{cos(\alpha)}{(sin(\alpha))^2}$$

$$D’ =-\frac{(3x^2-5x+5)\sqrt{(x+5)}}{2(x-5)(x^2+5)^{\frac{3}{2}}sin(\alpha)sin(\beta)}+\frac{1}{2\sqrt{(x^2-25)(x^3+5)} sin(\alpha)sin(\beta)}- \frac{(2x-5)\sqrt{x+5}}{2\sqrt{(x-5)(x^2+5)(x^2-5x+25)}} \frac{cos(\beta)}{(sin(\beta))^2sin(\alpha)} –\frac{(x+5)(2x^3+15x^2-5)}{2\sqrt{(x-5)(x^2+5)(x^3+5)}}\frac{cos(\alpha)}{(sin(\alpha))^2sin(\beta}$$ See the “degrees” in $N’$ and $D’$.

In $N’$ the second term has “degree” $-\frac 12$ and all of the other ones has “degree” $-\frac 32$. Hence $N’\to 0$ because the five terms tend to zero.

In $D’$ the first term has“degree” $-\frac 32$; the second one has “degree” $-\frac 52$, the third one has "degree" $-1$ and the fourth one has “degree” $1$.

Hence $D’\to\infty$ because the fourth term tends to $\infty$.

So we have got the $\frac{0}{\infty}$ shape that ends the proof.

Piquito
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  • Thank you @Ataulfo but doesn't $\frac{sin(z)}{z}{\to0}$ as x approaches infinity? And I am still confused about the lower case a,b and r. What exactly does this represent? – Arbuja Jul 14 '15 at 12:37
  • @Arbuja That limit $\frac{sin(z)}{z}{\to1}$ is well known (at the beginning for $z=x$; after for $z$ equal to "a good function” like you propose here). The A, B, C, D, a, b, r, are just elementary algebraic symbols showing identities to be used. – Piquito Jul 14 '15 at 12:56
  • So then are you taking the limit as x approaches zero or infinity? Because I found online the answer is zero as x approaches infinity. Please check even though I'm probably wrong. – Arbuja Jul 14 '15 at 13:11
  • @Arbuja You are right! I had a lapsus with the limit $\frac{sin(z)}{z}{\to1}$ (I have used many times "a good function" for this limit instead of x when I was student) . Sorry. I will try another answer to you (if I can, of course) – Piquito Jul 14 '15 at 14:11
  • Thanks @Atauflo for your answer. – Arbuja Jul 15 '15 at 10:47
  • @Arbuja: You are welcome.I was not sure I can get it. I am going to read your solution, so you have three instead of just one. This is always interesting. – Piquito Jul 15 '15 at 13:23
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One possibility is to factor the polynomials away from the sine function.

As $\lim_{x\to\infty}\left(\frac{x^2+5}{x+5}\right)^{1/2}-\left({x-5}\right)^{1/2}=0$

Since they are nearly equal to eachother we can factor them both out. You could take out the polynomial.

$$\lim_{x\to\infty}\left(\frac{x^2+5}{x+5}\right)^{1/2}\sin{\left(\left(\frac{x^3+5}{x+5}\right)^{1/2}\right)}-\left({x-5}\right)^{1/2}\sin{\left(\left({x^2-5x+25}\right)^{1/2}\right)}\approx$$

$$\lim_{x\to\infty}\left({x-5}\right)^{1/2}\left(\sin{\left(\left(\frac{x^3+5}{x+5}\right)^{1/2}\right)}-\sin{\left(\left({x^2-5x+25}\right)^{1/2}\right)}\right)$$

Now the mean value theorem can be used since the sine functions.

$$\lim_{x\to\infty}\left({x-5}\right)^{1/2}\cos(c)\left|\left(\frac{x^3+5}{x+5}\right)^{1/2}-\left({x^2-5x+25}\right)^{1/2}\right|$$

We can multiply $(x-5)$ into the absolute value...

$$\lim_{x\to\infty}\cos(c)\left|\left({x-5}\right)^{1/2}\left(\frac{x^3+5}{x+5}\right)^{1/2}-\left({x^2-5x+25}\right)^{1/2}\left({x-5}\right)^{1/2}\right|$$

From multiplication we get...

$$\lim_{x\to\infty}\cos(c)\left|\sqrt{\frac{x^4-5x^3+5x-25}{x+5}}-\sqrt{x^3-10x^2+50x-125}\right|$$

I could use l'hospitals theorem but it takes so long. Instead I would rather take the laurent series at $x=\infty$ since the radius of convergence is limited.

I first had to factor out both functions so that the terms of the laurent series ... $$\cos(c)\lim_{x\to\infty}\left|\sqrt{\frac{x^4-5x^3+5x-25}{x+5}}-\sqrt{x^3-10x^2+50x-125}\right|$$

$$\cos(c)\lim_{x\to\infty}\left|\sqrt{\frac{\left(\frac{1}{x}\right)^4-5\left(\frac{1}{x}\right)^3+5\left(\frac{1}{x}\right)-25}{\frac{1}{x}+5}}-\sqrt{\left(\frac{1}{x}\right)^3-10\left(\frac{1}{x}\right)^2+50\left(\frac{1}{x}\right)-125}\right|$$

$$\cos(c)\lim_{x\to\infty}\left|\sqrt{\frac{1}{x^{3}}}\sqrt{\frac{1-5x+5x^3-25x^4}{5x+1}}-\sqrt{\frac{1}{x^3}}\sqrt{1-10x+50x^2-125x^3}\right|$$

Taking the taylor series and substituting $\frac{1}{x}$ into x.

$$\cos(c)\lim_{x\to\infty}\left|\sqrt{{x^{3}}}\left(1-\frac{5}{x}- \frac{25}{2x^2}-\frac{60}{x^3}+O\left(\frac{1}{x^4}\right)\right)-\sqrt{{x^3}}\left(1-\frac{5}{x}+\frac{25}{x^2}+\frac{125}{x^3}+O\left(\frac{1}{x^4}\right)\right)\right|$$ Simplify....

$$\cos(c)\lim_{x\to\infty}\left|\left({x}^{3/2}-{5}{x^{1/2}}- \frac{25}{2x^{1/2}}-\frac{60}{x^{3/2}}+{x^{3/2}}O\left(\frac{1}{x^4}\right)\right)-\left(x^{3/2}-{5}{x^{1/2}}+\frac{25}{x^{1/2}}+\frac{125}{x^{3/2}}+x^{3/2}O\left(\frac{1}{x^4}\right)\right)\right|$$

The result is 0... I think I'm right but if not please correct me.

Arbuja
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  • Something hard however seem right, and meritorious for effort. Note the other two answers are based on two previous elementary transformation of ab-cd. Maybe there is another that can be used. Regards. – Piquito Jul 15 '15 at 14:32