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I have an exam next week and i really need some help with this task...

i have a picture

enter image description here

which i edited to show you the part that i can't understand.

  • The task is to prove that A is the limit of the sequence, using the definition.

I understand that he increased its value overall to show me that the left side is smaller than the right side.

What i don't know, is what's the method to increase/decrease a value of a sequence? I clearly can't add (or substract or whatever) any x to the numerator and any y to the denominator, there must be some rule...

And second, why did he have do that in the first place? Why did he check whether the right side is bigger than the left side? Why did he took a bigger number to do check all of this? Why couldn't he just skip that step??

thank you!

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1 Answer 1

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Essentially you are increasing the numerator and reducing the denominator, which is fine so long as everything stays positive.

If $x,y$ are both positive and $a \ge 0$ and $0 \lt b \lt y$ then you have

  • $\dfrac{x}{y} \le \dfrac{x+a}{y}$ and
  • $\dfrac{x}{y} \lt \dfrac{x}{y-b}$, so
  • $\dfrac{x}{y} \lt \dfrac{x+a}{y-b}$

Here $x=4n^2+3n$, $y=2n^4-n^2+5$, $a=3n^2-3n$, $b=n^4-n^2+5$ and presumably $n\ge 1$

The aim is to get the numerator and denominator into simpler forms, here $\dfrac{7n^2}{n^4}$ so you can then look at $\dfrac{7}{n^2}$

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  • $\begingroup$ But why did he need to create inequality? Why couldn't he just go along with equalities? I'm sorry, but for some reason i can't see why that move was necessary... Also, using your example, whats the method of choosing 'a' and 'b'? $\endgroup$
    – John Dale
    Oct 9, 2016 at 13:19
  • $\begingroup$ Equaility would have worked too and makes little difference. The choice of $a$ and $b$ here is to get the numerator and denominator into the simpler forms $cx^d$ so that dividing one by the other is easy $\endgroup$
    – Henry
    Oct 9, 2016 at 16:23

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