$$\lim_{x\to\infty}\left(\left(\frac{x^2+5}{x+5}\right)^{3.7}+\left(\frac{x^3+5}{x+5}\right)^{1.6}\right)^{20/37}-\left(\left(x-5\right)^{3.7}+(x^2-5x+25)^{1.6}\right)^{20/37}=60$$
It is possible to solve this limit using l'hospitals rule but I was asked to solve this using laurent series.
So far I tried working inside the parenthesis...
$$\left(\left(\frac{\frac{1}{x^2}+5}{\frac{1}{x}+5}\right)^{3.7}+\left(\frac{\frac{1}{x^3}+5}{\frac{1}{x}+5}\right)^{1.6}\right)^{20/37}-\left(\left(\frac{1}{x}-5\right)^{3.7}+\left(\frac{1}{x^2}-\frac{5}{x}+25\right)^{1.6}\right)^{20/37}$$
Then I used multiplication...
$$\left(\left(\frac{x^2}{x^2}\frac{\frac{1}{x^2}+5}{\frac{1}{x}+5}\right)^{3.7}+\left(\frac{x^3}{x^3}\frac{\frac{1}{x^3}+5}{\frac{1}{x}+5}\right)^{1.6}\right)^{20/37}-\left(\left(\frac{1-5x}{x}\right)^{3.7}+\left(\frac{1-5x+25x^2}{x^2}\right)^{1.6}\right)^{20/37}$$ Simplified..... $$\left(\left(\frac{1+5x^2}{x+5x^2}\right)^{3.7}+\left(\frac{1+5x^3}{x^2+5x^3}\right)^{1.6}\right)^{20/37}-\left(\left(\frac{1}{x}{1-5x}\right)^{3.7}+\left(\frac{1}{x^2}{1-5x+25x^2}\right)^{1.6}\right)^{20/37}$$ So I take the series at x=0 and substituted 1/x. $$\left(\left(x^{3.7}\right)\left(1-18.5\frac{1}{x}+235.875\frac{1}{x^2}-2704.31\frac{1}{x^3}+O\left(\frac{1}{x^4}\right)\right)+\left(x^{1.6}\right)^{2}\left(1-8\frac{1}{x}-304\frac{1}{x^3}+O\frac{1}{x^4}\right)\right)^{20/37}-\left(\left({x^2}\right)^{1.6}\left(1-8\frac{1}{x}+52\frac{1}{x^2}-112\frac{1}{x^3}+O\left(\frac{1}{x^4}\right)\right)+\left(x^{3.7}\right)\left(1-18.5\frac{1}{x}+124.875\frac{1}{x^2}-358.813\frac{1}{x^3}+O\left(\frac{1}{x^4}\right)\right)\right)^{\frac{20}{37}}$$
From there I am not sure what else to do. Am I solving this limit correctly and if not what are the steps to solving this?
Clearly the point of this exercise is that the two terms between large brackets have a very similar series expansion. As a consequence if one evaluates their difference, in the limit of $x$ to infinity, a lot of cancellation takes place.
The task therefore is to identify the lowest $x^n$ term in both expansions for which the pre-factors are not identical. The difference between these lowest non-identical pre-factors will determine the limiting behaviour of the overall expression for $x$ to infinity.
So let us examine the two terms between large brackets and identify their leading contributions.
Both terms have these two zero order contributions: $x^{3.7}$ and $x^{3.2}$
Both terms have these two first order contributions: $-18.5 x^{2.7}$ and $-16 x^{2.2}$.
The first term has a second order contribution $(5*3.7+12.5*3.7*4.7) x^{1.7}$, whereas the second term has instead the contribution $(12.5*3.7*2.7) x^{1.7}.$
Now we can perform the final evaluation of the requested limit. We must first factor out the leading term $x^{3.7}$. Thus we obtain:
$$x^{3.7*20/37}\{1 + ....... + (5*3.7+12.5*3.7*4.7) x^{-2}\}^{20/37} - x^{3.7*20/37}\{1 + ....... + (12.5*3.7*2.7) x^{-2}\}^{20/37}$$
$$= x^2 \{1 - 1 + (20/37)(5*3.7+12.5*3.7*4.7-12.5*3.7*2.7) x^{-2}\}$$
$$=2(5 + 25) = 60$$