lim of infinity sqrt9x to the 6th power -x all over xsquared +1 I got 3 to be the limit. If thats right, im not sure how I got the answer
\[\lim_{x\rightarrow \infty}\frac{\sqrt{9x^6-1}}{x^2+1}\]
if so, should be infinity
no I think the -1 is outside the square root
\[\lim_{x\rightarrow \infty}\frac{\sqrt{9x^6-x}}{x^2+1}\]
then it simplifies to (3x^3 - 1)/(x^2 + 1)
Then divide top and bottom by x^3 and you are left with 3/1
oh no!
numerator is a cube, denominator is a square plus how does is simplify?
nay is right
aahh yeas you're right
again you do this with your eyeballs. ignore everything but highest power. the numerator is not a polynomial, but \[\sqrt{x^6}=x^3\] so think "degree 3" denominator is a polynomial of degree 2 bigger numerator means bigger number, so no limit (or it is infinity)
its square root of 9x to the 6th power - x all over x cubed plus 1
ooooooooooooooooooh x cubed!!!!
hmmmm not to give you a hard time... but you wrote squared!
then you are right good work
haha yeah I think I solved it with x^3...yeah good job calculus.
good work to nayeaddo too
so then a problem like x+x cubed+x to the 5th power all over 1-x squared +x to the fourth power would be infinity???
lol @ satellite73
can you maybe write it using the equation tool?
\[\lim_{x\rightarrow \infty}\frac{x+x^3+x^5}{x-x^2+x^4}\] ?
that is infinity because the degree of the numerator is higher than the denominator
think about what it would be like if x were 1,000,000 = 10^6
\[(x + x^3 + x^5)\]/ (1 - x^2 + x^4)
but solutions is the same regardless.
or even ten for that matter. suppose you made x = 10 you would have \[\frac{101010}{10101}\]
numerator is bigger than denominator by a factor of ten. this gets huge as x gets huge
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