Question

Find the limit as x approaches negative infinity of (6/x)-(x/4)

Answer 1

@freckles

Answer 2

hint:
\[\frac{6}{x}-\frac{x}{4}=\frac{24-x^2}{4x} =\frac{\frac{24}{x}-\frac{x^2}{x}}{\frac{4x}{x}}=\frac{\frac{24}{x}-x}{4}\]

Answer 3

DNE?

Answer 4

or +ve infinity

Answer 5

While I do think you can say the limit dne, I think you could actually be more descriptive and say yes + inf since we would have
\[\frac{0-(-\inf)}{4}=\frac{0+\inf}{4}=\frac{\inf}{4}=\inf\]
And I hope you know what I just wrote is not formal at all
like you can't actually plug in infinity as if it were a number :p

Answer 6

yep

Answer 7

I understand :)

Answer 8

cool stuff :)

Answer 9

how good are u with physics?

Answer 10

@freckles

Answer 11

not very good
i actually had no physics class experience
not even one class :p

Answer 12

k

Answer 13

I have one more question

Answer 14

Find the limit as x approaches - infinity of x/(SQRT(X^2-x))

Answer 15

i think it will give infinity

Answer 16

you could divide both top and bottom by sqrt(x^2)

Answer 17

but you want to use that sqrt(x^2)=-x since x->-inf

Answer 18

dats -1?

Answer 19

\[\frac{\frac{x}{\sqrt{x^2}}}{\sqrt{1-\frac{1}{x}}}\]

Answer 20

actually yes
you would have -1/sqrt(1-0)
very nice

Answer 21

but can I do it like dis?