Answer:
see attached
Step-by-step explanation:
The limit as x gets large is the ratio of the highest-degree terms. In most cases, the limit can be found by evaluating that ratio. Where an absolute value is involved, the absolute value of the highest-degree term is used.
If the ratio gives x to a positive power, the limit does not exist. If the ratio gives x to a negative power, the limit is zero.
The arrangement of functions according to the given condition
[tex]m(x)=\frac{4x^{2}-6 }{1-4x^{2} }[/tex]
[tex]h(x)=\frac{x^{3} -x^{2} +4}{1-3x^{2} }[/tex]
[tex]k(x)=\frac{5x+1000}{x^{2} }[/tex]
[tex]i(x)=\frac{x-1}{|1-4x| }[/tex]
[tex]g(x)=\frac{|4x-1|}{x-4}[/tex]
[tex]l(x)=\frac{5x^{2} -4}{x^{2} +1}[/tex]
[tex]f(x)=\frac{x^{2} -1000}{x-5}[/tex]
[tex]j(x)=\frac{x^{2}-1 }{|7x-1|}[/tex]
A limit is the value that a function approaches as the input approaches some value.
According to the given question
[tex]l(x)=\frac{5x^{2} -4}{x^{2} +1}[/tex]
⇒[tex]\lim_{nx\to \infty} \frac{5x^{2} -1}{x^{2} +1}[/tex]
⇒[tex]\lim_{x \to \infty} \frac{x^{2} }{x^{2} } \frac{5-\frac{1}{x^{2} } }{1+\frac{1}{x^{2} } }[/tex]
= 5 ([tex]\frac{1}{x^{2} } = 0[/tex] ,as x tends to infinity [tex]\frac{1}{x^{2} }[/tex] tends to 0)
[tex]i(x)=\frac{x-1}{|1-4x|}[/tex]
⇒[tex]\lim_{x \to \infty} \frac{x-1}{|1-4x|}[/tex] = [tex]\lim_{x \to \infty} \frac{x}{x} \frac{1-\frac{1}{x} }{|\frac{-1}{4}+\frac{1}{x} | }[/tex] =[tex]\frac{1}{\frac{1}{4} }[/tex] =[tex]\frac{1}{4}[/tex]
As x tends to infinity 1/x tends to 0, and |[tex]\frac{-1}{4}[/tex]| gives 1/4
[tex]f(x)= \frac{x^{2} -1000}{x--5}[/tex]
⇒[tex]\lim_{x \to \infty} \frac{x^{2} -1000}{x-5}[/tex]= [tex]\lim_{x \to \infty} \frac{x^{2} }{x} \frac{1-\frac{1000}{x^{2} } }{1-\frac{5}{x} }[/tex]= [tex]\lim_{x \to \infty} x\frac{1-\frac{1000}{x^{2} } }{1-\frac{5}{x} }[/tex] ⇒ limit doesn't exist.
[tex]m(x)=\frac{4x^{2}-6 }{1-4x^{2} }[/tex]
⇒[tex]\lim_{x\to \infty} \frac{4x^{2} -6}{1-4x^{2} }[/tex]=[tex]\lim_{x\to \infty} \frac{x^{2} }{x^{2} } \frac{4-\frac{6}{x^{2} } }{\frac{1}{x^{2} } -4}[/tex] [tex]= \lim_{n \to \infty} \frac{4}{-4}[/tex] = -1
As x tends to infinity [tex]\frac{1}{x^{2} }[/tex] tends to 0.
[tex]g(x)=\frac{|4x-1|}{x-4}[/tex]
⇒[tex]\lim_{x\to \infty} \frac{|4x-1|}{x-4}[/tex] = [tex]\lim_{x \to \infty} \frac{|x|}{x} \frac{4-\frac{1}{x} }{1 -\frac{4}{x} } }[/tex] = 4
as x tends to infinity 1/x tends to 0
and |x|=x ⇒[tex]\frac{|x|}{x}=1[/tex]
[tex]h(x)=\frac{x^{3}-x^{2} +4 }{1-3x^{3} }[/tex][tex]\lim_{x \to \infty} \frac{x^{3} -x^{2} +4}{1-3x^{3} }[/tex][tex]= \lim_{x \to \infty} \frac{x^{3} }{x^{3} } \frac{1-\frac{1}{x}+\frac{4}{x^{3} } }{\frac{1}{x^{3} -3} }[/tex] = [tex]\frac{1}{-3}[/tex] =[tex]-\frac{1}{3}[/tex]
[tex]k(x)=\frac{5x+1000}{x^{2} }[/tex]
[tex]\lim_{x \to \infty} \frac{5x+1000}{x^{2} }[/tex] = [tex]\lim_{x \to \infty} \frac{x}{x} \frac{5+\frac{1000}{x} }{x}[/tex] =0
As x tends to infinity 1/x tends to 0
[tex]j(x)= \frac{x^{2}-1 }{|7x-1|}[/tex]
[tex]\lim_{x \to \infty} \frac{x^{2}-1 }{|7x-1|}[/tex] = [tex]\lim_{x \to \infty} \frac{x}{|x|}\frac{x-\frac{1}{x} }{|7-\frac{1}{x}| }[/tex] = [tex]\lim_{x \to \infty} 7x[/tex] = limit doesn't exist.
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