## ksaimouli 4 years ago find the domain of the function f.use limits to describe the behaviour of f at valus of x not in its domain 1/x+3

1. ksaimouli

plz explain me

2. slaaibak

For what value of x will you divide by 0?

3. ksaimouli

?

4. ksaimouli

$1\div x+1$

5. ksaimouli

|dw:1327444420449:dw|

6. slaaibak

Cool. $f(x) = {1 \over x+3}$ Where do you think will this function have a discontinuity? Aka, for what values of x is this function undefined?

7. ksaimouli

-3

8. slaaibak

Yes. So, do you agree this function is defined for every value of x, except for -3?

9. ksaimouli

yup

10. ksaimouli

so ow to write end bhaviour

11. ksaimouli

i am stuck with that

12. slaaibak

Then we can agree that the domain of the function is: $(-\infty,-3) \cup (-3, \infty)$

13. slaaibak

That means x goes from -infinity to JUST before three, meaning every value smaller than -3. and then it unites with every value larger than -3 to infinity

14. ksaimouli

thx slaaibak

15. ksaimouli

thank u ver much

16. slaaibak

For the limits: $\lim_{x \rightarrow -3^-} { 1 \over x + 3} = -\infty$ similarly $\lim_{x \rightarrow -3^+} { 1 \over x + 3} = \infty$ and lastly $\lim_{x \rightarrow -3} { 1 \over x + 3} = Does NotExist$

17. ksaimouli

ok got u but what happens in this case |dw:1327445191194:dw|

18. ksaimouli

i got it -2 right

19. slaaibak

Whichever x makes the denominator zero would make the undefined for that x. $(-\infty, -2) \cup (-2,2) \cup (2, \infty)$ That's the domain

20. slaaibak

if x^2 - 4 = 0 x = -2 or x = 2