Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

with your eyeballs

Answer 5

so sqrt3/-2?

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

??

Answer 10

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

Answer 11

ohh..so its - infinit?

Answer 12

right it goes to -inf

Answer 13

dunno but that looks like what you wrote yes?

Answer 14

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

Answer 15

i think so?

Answer 16

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

Answer 17

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

Answer 18

i don't agree

Answer 19

hello myininaya!

Answer 20

oops i do agree

Answer 21

well you are probably right and i am wrong as usual

Answer 22

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

Answer 23

oh then if you agree i disagree! lol

Answer 24

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

Answer 25

forget that noise (no offence)

Answer 26

i will write this out again in prettiness okay

Answer 27

okay:)

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

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

Answer 38

what about the power to 2?

Answer 39

not true satellite

Answer 40

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

Answer 41

what is not true?

Answer 42

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

Answer 43

way smarter

Answer 44

both of u r smart! im not

Answer 45

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

Answer 46

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

Answer 47

!!! no bike rides for you!

Answer 48

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

Answer 49

then you thought correctly!

Answer 50

didnt u say the opp of tht previosuly?

Answer 51

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

Answer 52

oops i didn't see you change the signs sorry

Answer 53

ohh right..4got abt tht negative

Answer 54

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

Answer 55

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

Answer 56

so the ans is -3/2?

Answer 57

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

Answer 58

ohhok...i think i got it

Answer 59

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

Answer 60

alrigty:) thanks a ton!

Answer 61

yw

Answer 62

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

Answer 63

jk

Answer 64

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

Answer 65

haha

Answer 66

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

Answer 67

haha thts funny

Answer 68

woof