Question

Absolute Value question:

So if I get sometime like lx-1l

And they ask for lim -infinity or lim 5^+

So for the lim -infinity would the function equal to just (x-1) and for the lim 5^+ the function will equal to -(x-1)?

Thanks,

Answer 1

you will get infinity in either case. do you have a specific example?

Answer 2

Like if lim x-> infinity , 3x-l2x-3l/5x+2 to get rid of the absolute value will I just put it in brackets and add a - so like 3x--(2x-3)/5x+2?

Answer 3

since x is going to infinity you can assume it is greater than 3/2 and so |2x-3|=2x-3

Answer 4

you would therefore be taking
\[\lim_{x\rightarrow 0}\frac{3x-(2x-3)}{5x+2}=\frac{x+3}{5x+2}\]

Answer 5

sorry i meant limit as x goes to infinity, not zero

Answer 6

and answer would be 1/5

Answer 7

Ohhh, so then if its asking for lim x-> -infinity you can then assume that x will be less then 3/2 so then I'll add a - sign so it would be 3x--(2x-3)?

Answer 8

I get it now thanks, one last question if you don't mind tho,
if the question is lim x-> 2 , 1/ lx-2l what would happen in that case? Thanks!

Answer 9

you need to look from left and right since |x-2| will be different in both cases
\[|x-2|=x-2 \text{ if } x-2>0 \text{ (or } x>2 )\]
\[|x-2|=-(x-2) \text{ if } x-2<0 \text{ (or } x<2 )\]

\[\lim_{x \rightarrow 2^-}\frac{1}{-(x-2)}=+\infty\]
\[\lim_{x \rightarrow 2^+}\frac{1}{x-2}=+ \infty\]

Answer 10

Ohh I get it, thanks a lot I appreciate it :D

Answer 11

satellite wrote what i think he did not mean to write

Answer 12

\[\lim_{x\rightarrow 0}\frac{3x-(2x-3)}{5x+2}=\lim_{x \rightarrow 0}\frac{x+3}{5x+2} =\frac{3}{2}\]
\[\lim_{x\rightarrow \infty}\frac{3x-(2x-3)}{5x+2}=\lim_{x\rightarrow \infty} \frac{x+3}{5x+2}=\frac{1}{5}\]

Answer 13

Oh ic, thats makes sense thanks, though just one question if you dont mind, for the left side where it says lim x-> 2^- 1/-(x-2) = infinity, how would that = to infinity?

Answer 14

\[\frac{1}{-(1.999-2)}=\frac{1}{-(-.001)}=\frac{1}{.001}=1000\]

Answer 15

1.999 is to the left of 2

Answer 16

OH, yeah I get it now, thanks!