Help with this limit?

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?

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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.

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
$$
\begin{align}
&\left|\left(\frac{x^2+5}{x+5}\right)^{1/2}\sin\left(\left(\frac{x^3+5}{x+5}\right)^{1/2}\right)-\left(\frac{x^2-25}{x+5}\right)^{1/2}\sin\left(\left(\frac{x^3+125}{x+5}\right)^{1/2}\right)\right|\\
&\le\left(\frac{x^2+5}{x+5}\right)^{1/2}\frac{60}{x^2}+1\cdot\frac{15}{(x-5)^{3/2}}\\[12pt]
&\to0\tag{4}
\end{align}
$$

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.