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pasta Group Title

find the lim (x+1/x) as x tends to 0.and sketch a graph to support for answer.

  • 2 years ago
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  1. 2le Group Title
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    \[lim_{x \rightarrow 0}(x+\frac{1}{x})=lim_{x \rightarrow 0}(\frac{x^2+1}{x})\] From here you can see that the function gets larger and larger as x approaches 0 from the left and the right.|dw:1349841452926:dw| also, if you look at \[lim_{x \rightarrow \infty}(\frac{x^2+1}{x})=2\] \[lim_{x \rightarrow -\infty}(\frac{x^2+1}{x})=2\]from l'Hopital's rule

    • 2 years ago
  2. pasta Group Title
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    as x tends to infinity the function is undefined as far as i know

    • 2 years ago
  3. adi171 Group Title
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    use l'hopital rule

    • 2 years ago
  4. adi171 Group Title
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    |dw:1350219305254:dw|

    • 2 years ago
  5. adi171 Group Title
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    then the x^2 cancels out.....the eq left is 1/x^2...... now when x approaches 0 the eq. approaches infinity,...!!1 so infinity is the ans

    • 2 years ago
  6. adi171 Group Title
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    graph is this and its correct

    • 2 years ago
  7. adi171 Group Title
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    |dw:1350219633729:dw|

    • 2 years ago
  8. adi171 Group Title
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    range of the grph is (-infinity , -2] union [2,infinity)

    • 2 years ago
  9. pasta Group Title
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    @adi171 there is no way i can use l 'HOpitals rule becoz the function gives 1/0 not 0/0 OR infnty/infnty.

    • 2 years ago
  10. pasta Group Title
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    when i use the rule my derivative is 2x which gives the answer 0.HOW ARE YOU differentiating 2x/1 which is actualy the derivative .and it does not work for the rule on this case

    • 2 years ago
  11. pasta Group Title
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    you are finding the minimum and maximum points of this graph. as x tends to inifinity the function tends to infinity .and AS X tends to 0 the function is undefined(see the attached graph)

    • 2 years ago
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