Question

lim of x approaching negative infinity (2x+3)/ (sqrt x^2+x+1) is...

Answer 1

my guess is \(-2\)

Answer 2

want me to check it?

Answer 3

how would you get that answer?

Answer 4

it could be wrong, because i did it with my eyeballs
ignore everthing but the highest term
\[\frac{2x}{\sqrt{x^2}}=\frac{x}{|x|}\] and since \(x<0\) you get \(-2\)

Answer 5

typo there, should have been \[\frac{2x}{\sqrt{x^2}}=\frac{2x}{|x|}=-2\]

Answer 6

Ok thank you!

Answer 7

hold on maybe i am wrong
lets check it

Answer 8

Ok

Answer 9

yeah its right
http://www.wolframalpha.com/input/?i=limit+x+to+-+infty+%282x%2B3%29%2F%28sqrt%28x^2%2Bx%2B1%29%29

Answer 10

you can also do some tedious calculation where you divide top and bottom by \(x\) but that is annoying and unnecessary

Answer 11

ok thank you for your help!