lim x->-infinti of sqrt (3x^2+6) / 5-2x

\[\lim_{x \rightarrow -\infty} \sqrt{3x ^{2+6}} \div 5-2x\]

\[\frac{\sqrt{3}}{2}\]

\[\frac{\sqrt{3x^2+6}}{5-2x}\]

with your eyeballs

so sqrt3/-2?

no because you are going to \[-\infty\] yes?

\[\frac{\sqrt{3x^2+6}}{5-2x}*\frac{\frac{1}{x}}{\frac{1}{x}}\] =\[\frac{\sqrt{\frac{3x^2}{x}+\frac{6}{x}}}{{\frac{5}{x}-\frac{2x}{x}}}\]

if you were going to \[\infty\] you would be right. but you are going to \[-\infty\] so both top and bottom are positive

??

\[\frac{\sqrt{3x+\frac{6}{x}}}{\frac{5}{x}-2}\]

ohh..so its - infinit?

right it goes to -inf

dunno but that looks like what you wrote yes?

lim x->-infinti of sqrt (3x^2+6) / 5-2x

i think so?

of x->-inf didn't see the negative

ok in that case it is \[\frac{\sqrt{3}}{2}\] fer sure

i don't agree

hello myininaya!

oops i do agree

well you are probably right and i am wrong as usual

lol so step by step..the powers dont match..so i thought it wud jus b 0 or doesnt exisit?

oh then if you agree i disagree! lol

i divided top by wrong thing it is suppose to be sqrt(x^2)

forget that noise (no offence)

i will write this out again in prettiness okay

okay:)

look. denominator is poly of degree 1 yes? numerator is not a poly but for limit purposes it might as well be. nothing matter but highest degree

forget everything after the \[\sqrt{3x^2}=\sqrt{3}x\]

\[\frac{\sqrt{3x^2+6}}{5-2x}*\frac{\frac{1}{x}}{\frac{1}{x}}=\frac{\sqrt{\frac{3x^2+6}{x^2}}}{\frac{5}{x}-\frac{2x}{x}}\]

so you get \[\frac{\sqrt{3}x}{-2x}=\frac{\sqrt{3}}{-2}\]

\[\frac{\sqrt{3+\frac{6}{x^2}}}{\frac{5}{x}-2}\]

not that my esteemed colleague is wrong, it is just that you can do this with your eyes

6/x^2->0 5/x->0

but isnt it \[\sqrt{3x ^{2}}\]

anyway i should be quiet because myininaya is way smarter than i am

right and \[\sqrt{3x^2}=\sqrt{3}x\]\]

what about the power to 2?

not true satellite

\[\sqrt{x^2}=|x|\]

what is not true?

since x->-inf, f(x)->sqrt{3}/2 if x->inf, f(x)->-sqrt{3}/2 that im smater (its not true)

way smarter

both of u r smart! im not

but i did cross the 1000 mark! no one threw a party :(

i threw a party for you and i forgot to invite you sorry

!!! no bike rides for you!

since x->-inf, f(x)->-sqrt{3}/2 if x->inf, f(x)->sqrt{3}/2 i thought it was this?

then you thought correctly!

didnt u say the opp of tht previosuly?

you have a - sign in the denominator. so no sorry

oops i didn't see you change the signs sorry

ohh right..4got abt tht negative

denominator is 5 - 2x so as x -> infinity this goes to - infinity and vice versa

if x is negative and there is a negative sign in front of it that number would be positive right?

so the ans is -3/2?

i am taking my 1005 medals and going to bed. no \[\frac{\sqrt{3}}{2}\]

ohhok...i think i got it

good. don't forget the coefficient in the numerator is \[\sqrt{3}\] not 3

alrigty:) thanks a ton!

yw

sanhitha you forgot to ask him if he is indian im disappointed in you

jk

lol..dude! i only asked 2 ppl includin u

haha

actually... http://www.unc.edu/depts/jomc/academics/dri/idog.jpg

haha thts funny

woof

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