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Gnom [1K]
3 years ago
11

Limit as x approaches infinity: 2x/(3x²+5)

Mathematics
1 answer:
Nonamiya [84]3 years ago
5 0
\bf \lim\limits_{x\to \infty}~\cfrac{2x}{3x^2+5}\implies \cfrac{\lim\limits_{x\to \infty}~2x}{\lim\limits_{x\to \infty}~3x^2+5}

now, by traditional method, as "x" progresses towards the positive infinitity, it becomes 100, 10000, 10000000, 1000000000 and so on, and notice, the limit of the numerator becomes large.

BUT, notice the denominator, for the same values of "x", the denominator becomes larg"er" than the numerator on every iteration, ever becoming larger and larger, and yielding a fraction whose denominator is larger than the numerator.

as the denominator increases faster, since as the lingo goes, "reaches the limit faster than the numerator", the fraction becomes ever smaller an smaller ever going towards 0.

now, we could just use L'Hopital rule to check on that.

\bf \lim\limits_{x\to \infty}~\cfrac{2x}{3x^2+5}\stackrel{LH}{\implies }\lim\limits_{x\to \infty}~\cfrac{2}{6x}

notice those derivatives atop and bottom, the top is static, whilst the bottom is racing away to infinity, ever going towards 0.
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I need help with that
alukav5142 [94]
If i remember correctly.. it’s C, you cant have the X values repeating.
8 0
3 years ago
Read 2 more answers
A) how many ways are there to choose 10 players to take the field?
Kruka [31]

A. C(13, 10) = 13! = 13·12·11 = 13 · 2 · 11 = 286.C(13, 10) = 13! = 13·12·11 = 13 · 2 · 11 = 286.

B. P(13,10)= 13! =13! =13·12·11·10·9·8·7·6·5·4. (13−10)! 3!

C. f there is exactly one woman chosen, this is possible in C(10, 9)C(3, 1) =

10! 3!

9!1! 1!2! 10! 3!

8!2! 2!1! 10! 3!

7!3! 3!0!

= 10 · 3 = 30 ways; two women chosen — in C(10,8)C(3,2) =

= 45·3 = 135 ways; three women chosen — in C(10, 7)C(3, 3) =

= 10·9·8 ·1 = 120 ways. Altogether there are 30+135+120 = 285 1·2·3

<span>possible choices.</span><span> 
</span>

  


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3 years ago
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Step-by-step explanation:

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3 years ago
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