Question

Help, please. №249 http://s006.radikal.ru/i215/1410/9b/b4847e07c4f3.jpg

Answer 1

That looks really hard

Answer 2

is the answer "1" ??

Answer 3

yes

Answer 4

you just put \[x ^{100}\] in evidence out of the equation then you get
\[\lim_{x \rightarrow \infty} = \frac{ x ^{100}*(1 + \frac{ 1 }{ x ^{99} }... )}{ x ^{100} *( 1 + \frac{ 1 }{ x^{99} }....)}\]
You will get something like this.. anything that has the quociente X will aproximate zero and the \[\frac{ x ^{100} }{ x ^{100} } = 1\]
than you have in the end \[\lim_{x \rightarrow \infty} = \frac{ 1*(1) }{ 1*(1) }\]

Got it?

Answer 5

i don't understand why if you put x^100 you get this expression in the brackets

Answer 6

P(x)=an xn+a(n-1) x(n-1)+...+a2 x2+a1 x+a0
you just put the highest polinomium in evidence (Out of the ()).. and you get that equation

Answer 7

for example \[(x ^{100}+1) = x ^{100}*(1 + \frac{ 1 }{ x ^{100} })\]

Answer 8

you just have to know what are the "highest" power of x you get and put it out of the brackets.

Answer 9

okay, thank you!

Answer 10

did you really get it? I can try to explain in another way

Answer 11

i think i got it, but maybe you meant x^100 in the denominator? not x^infinity

Answer 12

yep.. you have x^100 on both denominator and numerator

Answer 13

then i got it, thanks!

Answer 14

the hint is, when you got these kind of question all you have to pay attention is to the highest powers.. like
http://www.somatematica.com.br/superior/limites/limite63.gif

Answer 15

i understand now, thanks again :)