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anonymous

  • one year ago

How would I solve this?

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  1. anonymous
    • one year ago
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  2. SolomonZelman
    • one year ago
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    \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~7^-}~\dfrac{1}{x-7}}\)

  3. SolomonZelman
    • one year ago
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    well, you know (I would assume) that when you are dealing with a function f(x)=1/(x-a) then, your asymptote is x=a

  4. SolomonZelman
    • one year ago
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    So, this way you should be able to identify the asymptote.

  5. anonymous
    • one year ago
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    So, B?

  6. SolomonZelman
    • one year ago
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    Now, you are asked to use a numerical approach (and make a table of values) for this limit. I will show you what to do, and you will complete the chart: keep in mind, your function is \(f(x)=\dfrac{1}{x-7}\) and in your limit, the x is appraching 7 from the left side, so we start from values that are smaller than 7 just a bit, and then approach 7 closer and closer "from the left". \(\large\color{black}{ \huge{ \begin{array}{| l | c | r |} \hline \scr~~~x~~ & \scr~~~ f(x) ~~~\\ \hline \scr~~~6~ & \scr ~ \\ \hline \scr~~~6.5~ & \scr ~ \\ \hline \scr~~~6.9~ & \scr ~ \\ \hline \scr~~~6.99~ & \scr ~ \\ \hline \scr~~~6.999~ & \scr ~ \\ \hline \end{array} } }\) and using this table you should be able to find the limit: \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~7~^-}~\dfrac{1}{x-7}}\)

  7. SolomonZelman
    • one year ago
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    u can use wolframalpha.com to calculate these values for the table. and these values don't need to be exactly those, just some values that closer and closer approach 7 "from the left".

  8. anonymous
    • one year ago
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    -1 -2 -10 -100 -1000

  9. SolomonZelman
    • one year ago
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    yes.... see where the limit is going as x approaches 7 closer and closer?

  10. anonymous
    • one year ago
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    -infinity

  11. SolomonZelman
    • one year ago
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    yes, this limit is going towards -∞

  12. SolomonZelman
    • one year ago
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    Any questions?

  13. anonymous
    • one year ago
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    Unless you're willing to help me with another problem, no.

  14. SolomonZelman
    • one year ago
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    Oh, ok. But in different post if you will. (i will see if I can make it there)

  15. anonymous
    • one year ago
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    Thank you.

  16. SolomonZelman
    • one year ago
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    anytime!

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