## anonymous one year ago The graph of a function f is given above.

1. anonymous

What is $\lim_{x \rightarrow 2}\frac{ x }{ f(x)+1 }$

2. dinamix

f(x) equal what mate

3. anonymous

|dw:1440258551985:dw|

4. Vocaloid

well, start by finding f(2)

5. anonymous

f(2)=-1

6. dinamix

its infinity this lim but dont know + or -

7. dinamix

its + infinity

8. anonymous

How do I know if it's positive or negative Infiniti?

9. amistre64

1 - (something smaller than 1) is a positive number; and x=2 is a positive number +/+ = +

10. amistre64

|dw:1440260968991:dw|

11. anonymous

I forgot to make the bottom of the graph negative oops :(

12. anonymous

|dw:1440261046439:dw|

13. amistre64

you see that for say x between 1 and 3, that the top of the setup is positive and that the bottome part: f(1,3) + 1 is just the same as: 1 - (a fraction) is positive

14. anonymous

How did you get f(1,3) +1 and 1-(a fraction)?

15. amistre64

i glossed over x=2 :) other than that you see that f(x) is a value between 0 and -1 right?

16. amistre64

-a + 1 = 1 - a

17. amistre64

and saying "a fraction" is lazy as well, since the values are both rational and irrational.

18. anonymous

Oh ok :) so f(1,3) means those are the x values that cross on 0 and that's why you add 1?

19. amistre64

f(x) + 1 is the bottom of your setup and if we just focus on the values between 1 and 3; f(1,3) just denotes all the values we can conceive of yes

20. amistre64

x = (1,3) includes x=2 f(1,3) includes all the values that x produces along the stated domain

21. anonymous

Ohh ok :)

22. anonymous

So do I solve for 1 first, and the 3?

23. amistre64

no, i was just making an observation the value of the bottom, when x=2, is 0 since f(2) = -1 now, the limit is defined if the value that is approached from the left and right of x=2 is the same. so we observe that for a small enough region around x=2, the top is always positive, and the bottom is always positive; therefore, we approach the same infinity (+inf). but in may classes, infinity is still classified as DNE

24. anonymous

Ohh ok I understand now :)

25. amistre64

good luck

26. anonymous

Thank you!!!!

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