Notice that:
[tex]\begin{gathered} \lim _{x\rightarrow\infty}\frac{2x^2(x-5)^2}{x^4}=\lim _{x\rightarrow\infty}\frac{2(x-5)^2}{x^2} \\ =\lim _{x\rightarrow\infty}\frac{2(x^2-10x+25)}{x^2} \\ =\lim _{x\rightarrow\infty}(\frac{2x^2}{x^2}-\frac{10x}{x^2}+\frac{25}{x^2}) \\ =\lim _{x\rightarrow\infty}(2-\frac{10}{x}+\frac{25}{x^2}) \\ =\lim _{x\rightarrow\infty}2-\lim _{x\rightarrow\infty}\frac{10}{x}+\lim _{x\rightarrow\infty}\frac{25}{x^2} \\ =2-0+0 \\ =2 \end{gathered}[/tex]
Which means that, for large values of x:
[tex]2x^2(x-5)^2\approx2x^4[/tex]
Since the function:
[tex]f(x)=2x^4[/tex]
is a 4th degree monomial, with positive coefficient, then it keeps growing as x grows. Then, we know that:
[tex]\lim _{x\rightarrow\infty}2x^2(x-5)^2=\infty[/tex]
Which means that the end behavior is such that f(x) approaches infinity as x approaches infinity.
On the other hand, for large negative values of x, the function is also positive. Then:
[tex]\lim _{x\rightarrow-\infty}2x^2(x-5)^2=\infty[/tex]
