Mathematics 25 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

$\frac{\sqrt{3}}{2}$

myininaya (myininaya):

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

OpenStudy (anonymous):

OpenStudy (anonymous):

so sqrt3/-2?

OpenStudy (anonymous):

no because you are going to $-\infty$ yes?

myininaya (myininaya):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

??

myininaya (myininaya):

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

OpenStudy (anonymous):

ohh..so its - infinit?

myininaya (myininaya):

right it goes to -inf

OpenStudy (anonymous):

dunno but that looks like what you wrote yes?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

i think so?

myininaya (myininaya):

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

OpenStudy (anonymous):

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

myininaya (myininaya):

i don't agree

OpenStudy (anonymous):

hello myininaya!

myininaya (myininaya):

oops i do agree

OpenStudy (anonymous):

well you are probably right and i am wrong as usual

OpenStudy (anonymous):

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

OpenStudy (anonymous):

oh then if you agree i disagree! lol

myininaya (myininaya):

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

OpenStudy (anonymous):

forget that noise (no offence)

myininaya (myininaya):

i will write this out again in prettiness okay

OpenStudy (anonymous):

okay:)

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

myininaya (myininaya):

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

OpenStudy (anonymous):

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

myininaya (myininaya):

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

OpenStudy (anonymous):

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

myininaya (myininaya):

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

OpenStudy (anonymous):

but isnt it $\sqrt{3x ^{2}}$

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

what about the power to 2?

myininaya (myininaya):

not true satellite

OpenStudy (anonymous):

$\sqrt{x^2}=|x|$

OpenStudy (anonymous):

what is not true?

myininaya (myininaya):

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

OpenStudy (anonymous):

way smarter

OpenStudy (anonymous):

both of u r smart! im not

OpenStudy (anonymous):

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

myininaya (myininaya):

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

OpenStudy (anonymous):

!!! no bike rides for you!

OpenStudy (anonymous):

myininaya (myininaya):

then you thought correctly!

OpenStudy (anonymous):

didnt u say the opp of tht previosuly?

OpenStudy (anonymous):

myininaya (myininaya):

oops i didn't see you change the signs sorry

OpenStudy (anonymous):

ohh right..4got abt tht negative

OpenStudy (anonymous):

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

myininaya (myininaya):

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

OpenStudy (anonymous):

so the ans is -3/2?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

ohhok...i think i got it

OpenStudy (anonymous):

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

OpenStudy (anonymous):

alrigty:) thanks a ton!

OpenStudy (anonymous):

yw

myininaya (myininaya):

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

myininaya (myininaya):

jk

OpenStudy (anonymous):

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

OpenStudy (anonymous):

haha

OpenStudy (anonymous):
OpenStudy (anonymous):

haha thts funny

OpenStudy (anonymous):

woof

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