Question

Use graphical and numerical evidence to conjecture a value for the indicated limit. lim x -> infinity sign (x^2)/(2^x)

Answer 1

@zepdrix can u help?

Answer 2

Hmm what is the `sign` for?

Answer 3

I read it as "infinity sign" -> \(\infty\)

Answer 4

the infinity sing yes like space showed

Answer 5

sign*

Answer 6

\[\Large \lim_{x \to \infty} \frac{x^2}{2^x}\]

Answer 7

so infinity sign with a negative on it yes space it that it

Answer 8

yea space that the problem

Answer 9

oh i see :) lol

Answer 10

My english is a bit flawed, but does it include now a minus sign or not? I haven't investigated this limit yet, but I am sure it makes a big difference, because 2^x is a super fast growing function compared to x^2, hence it will reach infinity much faster then the numerator.

Answer 11

no it doesn't include a minus sign

Answer 12

\[\Large \lim_{x \to \infty} \frac{x^2}{e^{\ln2 x}} = \frac{\infty}{\infty} \]
and then De L'hôpital.

\[\Large \lim_{x \to \infty} \frac{2x}{\ln2e^{\ln 2 x}}= \lim_{x\to \infty} \frac{2}{(\ln2)^2e^{\ln2x}} =\frac{2}{\infty}=0\]

Answer 13

so that how you conjecture the limit?

Answer 14

Doubt that this counts as numerical evidence, it's rather analytical, but as for numeric, the runtime (time complexity) of 2^x is of a much higher order than 2^x, therefore it reaches infinity much faster then the numerator and that yields to zero.

Answer 15

ok so how will i graph that information?

Answer 16

graphical evidence means to draw or compute a graph, since they mention that option I guess you have a graphical calculator? If not, wolframalpha can do that for you.

Answer 17

okay i'll try that. thanks

Answer 18

you're welcome

Answer 19

so look at my graph is it correct?

Answer 20

@Spacelimbus

Answer 21

http://www.wolframalpha.com/input/?i=graph+%28x%5E2%29%2F2%5Ex+from+x%3D-2+to+x%3D+10

Answer 22

ok i was wayy off.

Answer 23

it's a complex graph.