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?

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.
