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DaniilM [7]
2 years ago
10

PLEASE HELP

Mathematics
1 answer:
kolezko [41]2 years ago
6 0

Answer:

The arrangement of the functions for which the result is a non-infinite value and the limit exists in ascending order of their limit values as x tends to infinity is given below;

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

Step-by-step explanation:

Since the limit of all the function as x tends to infinity is a non-infinite value, the limit converges.

We must understand that lim of a/x as x tends to infinity is equivalent to zero where a is any constant.

Let's take the function one after the other.

For l(x) = 5x²-4/x²+1

Dividing through by highest power of x i.e x²

lim x-> \infty 5x²-4/x²+1

lim x-> \infty 5x²/x²-4/x²/x²/x²+1/x²

lim x->∞ 5-4/x²/1+1/x²

=5-4/∞/{1+1/∞²}

= 5-0/1+0

= 5/1

= 5

Hence lim x-> \infty 5x²-4/x²+1 = 5

Find the remaining solution in the attachment below.

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\bf \begin{array}{llll}
term&value\\
-----&-----\\
a_{10}&41\\
a_{11}&41+d\\
a_{12}&(41+d)+d\\
&41+2d\\
a_{13}&(41+2d)+d\\
&41+3d\\
a_{14}&(41+3d)+d\\
&41+4d\\
a_{15}&(41+4d)+d\\
&41+5d=61
\end{array}
\\\\\\
41+5d=61\implies 5d=20\implies d=\cfrac{20}{5}\implies \boxed{d=4}\\\\
-------------------------------\\\\

\bf n^{th}\textit{ term of an arithmetic sequence}\\\\
a_n=a_1+(n-1)d\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
d=\textit{common difference}\\
----------\\
d=4\\
n=10\\
a_{10}=41
\end{cases}
\\\\\\
41=a_1+(10-1)4\implies 41=a_1+36\implies \boxed{5=a_1}

thus

\bf n^{th}\textit{ term of an arithmetic sequence}\\\\
a_n=a_1+(n-1)d\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
d=\textit{common difference}\\
----------\\
d=4\\
n=3\\
a_{1}=5
\end{cases}
\\\\\\
a_3=a_1+(3-1)4\implies a_3=5+(3-1)4

and surely you know how much that is.
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2 years ago
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<h2>Answer:</h2>

Shown in the explanation

<h2>Step-by-step explanation:</h2>

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