Limits again

$\begin{aligned} \mathfrak L & = \lim_{x \to \infty} \frac {2((1+x)^5+(1-x)^5) ((1+x)^3 +(1-x)^3 )} {((1+x)^2 +(1-x)^2 )^3} \\ & = \lim_{x \to \infty} \frac{8(1+10x^2+5x^4)(1+3x^2)} {8 (1+ x^2) ^3} \\ & = \lim_{x \to \infty} \frac {1 +13x^2 +35x^4+15x^6}{1+3x^2 +3x^4+x^6} \quad \quad \small \color{#3D99F6} {\text {Divide up and down by } x^6} \\ & = \lim_{x \to \infty} \frac {x^{-6} +13x^{-4} +35x^{-2} +15}{x^{-6}+3x^{-4}+3x^{-2}+1} \\ & = \boxed{15} \end{aligned}$

The limit will depend only on the highest powers of x, so we don't need to expand the whole thing: It's easy to see that the $x^5$ and the $x^3$ terms above will cancel each other, so the highest-degree terms lead to: $Lim = \frac{2(10x^4)(3x^2)}{(2x^2)^3}$ Which is 15.

$\Rightarrow(1+x)^2+(1-x)^2=1+2x+x^2+1-2x+x^2=2x^2+2$

$\Rightarrow(1+x)^3+(1-x)^3=(1+x+1-x)[(1+x)^2-(1+x)(1-x)+(1-x)^2]=2(2x^2+2-1+x^2)=6x^2+2$

Since,

$[(1+x)^2+(1-x)^2][(1+x)^3+(1-x)^3]=(1+x)^5+(1-x)^5+[(1+x)(1-x)]^2(1+x+1-x)$

Thus,

$\Rightarrow(1+x)^5+(1-x)^5=(2x^2+2)(6x^2+2)-2(1-x^2)^2=12x^4+16x^2+4-2x^4+4x^2-2=10x^4+20x^2+2$

From Limit,

$\begin{aligned}\displaystyle\lim_{x\to\infty}\dfrac{2[(1+x)^5+(1-x)^5][(1+x)^3+(1-x)^3]}{[(1+x)^2+(1-x)^2]^3}=&\displaystyle\lim_{x\to\infty}\dfrac{2(10x^4+20x^2+2)(6x^2+2)}{(2x^2+2)^3}\\=&\displaystyle\lim_{x\to\infty}\dfrac{8(5x^4+10x^2+1)(3x^2+1)}{8(x^2+1)^3}\\=&\displaystyle\lim_{x\to\infty}\dfrac{15x^6+35x^4+13x^2+1}{x^6+3x^4+3x^2+1}\\=&\displaystyle\lim_{x\to\infty}\dfrac{15+\dfrac{35}{x^2}+\dfrac{13}{x^4}+\dfrac1{x^6}}{1+\dfrac3{x^2}+\dfrac3{x^4}+\dfrac1{x^6}}\\=&\boxed{15}\end{aligned}$