drag each tile to the correct box. not all tiles will be used. arrange the functions for which the result is…

Explanation:

Step1: Recall limit - rules for rational functions

For a rational function $y=\frac{a_nx^n + a_{n - 1}x^{n - 1}+\cdots+a_0}{b_mx^m + b_{m - 1}x^{m - 1}+\cdots+b_0}$, if $n=m$, $\lim_{x\rightarrow\infty}y=\frac{a_n}{b_m}$; if $n\lt m$, $\lim_{x\rightarrow\infty}y = 0$; if $n>m$, $\lim_{x\rightarrow\infty}y=\pm\infty$. For absolute - value functions, we consider the cases when $x$ is large.

Step2: Calculate $\lim_{x\rightarrow\infty}l(x)$

$l(x)=\frac{5x^2-4}{x^2 + 1}$, since the degrees of the numerator and denominator are the same ($n = m=2$), $\lim_{x\rightarrow\infty}l(x)=\lim_{x\rightarrow\infty}\frac{5x^2/x^2-4/x^2}{x^2/x^2 + 1/x^2}=\lim_{x\rightarrow\infty}\frac{5-\frac{4}{x^2}}{1+\frac{1}{x^2}} = 5$.

Step3: Calculate $\lim_{x\rightarrow\infty}i(x)$

$i(x)=\frac{x - 1}{|1 - 4x|}$. When $x\rightarrow\infty$, $|1 - 4x|=4x - 1$ (because for large $x$, $1-4x<0$). So $\lim_{x\rightarrow\infty}i(x)=\lim_{x\rightarrow\infty}\frac{x - 1}{4x - 1}=\lim_{x\rightarrow\infty}\frac{x/x-1/x}{4x/x - 1/x}=\frac{1 - 0}{4 - 0}=\frac{1}{4}$.

Step4: Calculate $\lim_{x\rightarrow\infty}f(x)$

$f(x)=\frac{x^2-1000}{x - 5}$, since the degree of the numerator ($n = 2$) is greater than the degree of the denominator ($m = 1$), $\lim_{x\rightarrow\infty}f(x)=\infty$, so we discard this function.

Step5: Calculate $\lim_{x\rightarrow\infty}m(x)$

$m(x)=-\frac{4x^2-6}{1 - 4x^2}$, since the degrees of the numerator and denominator are the same ($n = m = 2$), $\lim_{x\rightarrow\infty}m(x)=\lim_{x\rightarrow\infty}-\frac{4x^2/x^2-6/x^2}{1/x^2-4x^2/x^2}=\lim_{x\rightarrow\infty}-\frac{4-\frac{6}{x^2}}{\frac{1}{x^2}-4}=1$.

Step6: Calculate $\lim_{x\rightarrow\infty}g(x)$

$g(x)=\frac{|4x - 1|}{x - 4}$. When $x\rightarrow\infty$, $|4x - 1|=4x - 1$, so $\lim_{x\rightarrow\infty}g(x)=\lim_{x\rightarrow\infty}\frac{4x - 1}{x - 4}=\lim_{x\rightarrow\infty}\frac{4x/x-1/x}{x/x - 4/x}=4$.

Step7: Calculate $\lim_{x\rightarrow\infty}h(x)$

$h(x)=\frac{x^3-x^2 + 4}{1 - 3x^3}$, since the degrees of the numerator and denominator are the same ($n = m = 3$), $\lim_{x\rightarrow\infty}h(x)=\lim_{x\rightarrow\infty}\frac{x^3/x^3-x^2/x^3 + 4/x^3}{1/x^3-3x^3/x^3}=\lim_{x\rightarrow\infty}\frac{1-\frac{1}{x}+\frac{4}{x^3}}{\frac{1}{x^3}-3}=-\frac{1}{3}$.

Step8: Calculate $\lim_{x\rightarrow\infty}k(x)$

$k(x)=\frac{5x + 1000}{x^2}$, since the degree of the numerator ($n = 1$) is less than the degree of the denominator ($m = 2$), $\lim_{x\rightarrow\infty}k(x)=\lim_{x\rightarrow\infty}\frac{5x/x^2+1000/x^2}{x^2/x^2}=\lim_{x\rightarrow\infty}\frac{\frac{5}{x}+\frac{1000}{x^2}}{1}=0$.

Step9: Calculate $\lim_{x\rightarrow\infty}j(x)$

$j(x)=\frac{x^2-1}{|7x - 1|}$. When $x\rightarrow\infty$, $|7x - 1|=7x - 1$, and since the degree of the numerator ($n = 2$) is greater than the degree of the denominator ($m = 1$), $\lim_{x\rightarrow\infty}j(x)=\infty$, so we discard this function.

Step10: Arrange the functions in ascending order of their limits

The functions with non - infinite limits are $k(x),h(x),i(x),m(x),g(x),l(x)$. The limits are $\lim_{x\rightarrow\infty}k(x)=0$, $\lim_{x\rightarrow\infty}h(x)=-\frac{1}{3}$, $\lim_{x\rightarrow\infty}i(x)=\frac{1}{4}$, $\lim_{x\rightarrow\infty}m(x)=1$, $\lim_{x\rightarrow\infty}g(x)=4$, $\lim_{x\rightarrow\infty}l(x)=5$. The ascending order is $h(x),k(x),i(x),m(x),g(x),l(x)$.

Answer:

$h(x),k(x),i(x),m(x),g(x),l(x)$