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anonymous

  • one year ago

The graph of a function f is given above.

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  1. anonymous
    • one year ago
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    What is \[\lim_{x \rightarrow 2}\frac{ x }{ f(x)+1 }\]

  2. dinamix
    • one year ago
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    f(x) equal what mate

  3. anonymous
    • one year ago
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    |dw:1440258551985:dw|

  4. Vocaloid
    • one year ago
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    well, start by finding f(2)

  5. anonymous
    • one year ago
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    f(2)=-1

  6. dinamix
    • one year ago
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    its infinity this lim but dont know + or -

  7. dinamix
    • one year ago
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    its + infinity

  8. anonymous
    • one year ago
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    How do I know if it's positive or negative Infiniti?

  9. amistre64
    • one year ago
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    1 - (something smaller than 1) is a positive number; and x=2 is a positive number +/+ = +

  10. amistre64
    • one year ago
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    |dw:1440260968991:dw|

  11. anonymous
    • one year ago
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    I forgot to make the bottom of the graph negative oops :(

  12. anonymous
    • one year ago
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    |dw:1440261046439:dw|

  13. amistre64
    • one year ago
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    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
    • one year ago
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    How did you get f(1,3) +1 and 1-(a fraction)?

  15. amistre64
    • one year ago
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    i glossed over x=2 :) other than that you see that f(x) is a value between 0 and -1 right?

  16. amistre64
    • one year ago
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    -a + 1 = 1 - a

  17. amistre64
    • one year ago
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    and saying "a fraction" is lazy as well, since the values are both rational and irrational.

  18. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    x = (1,3) includes x=2 f(1,3) includes all the values that x produces along the stated domain

  21. anonymous
    • one year ago
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    Ohh ok :)

  22. anonymous
    • one year ago
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    So do I solve for 1 first, and the 3?

  23. amistre64
    • one year ago
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    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
    • one year ago
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    Ohh ok I understand now :)

  25. amistre64
    • one year ago
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    good luck

  26. anonymous
    • one year ago
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    Thank you!!!!

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