anonymous
  • anonymous
find the lim (x+1/x) as x tends to 0.and sketch a graph to support for answer.
MIT 18.01 Single Variable Calculus (OCW)
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anonymous
  • anonymous
find the lim (x+1/x) as x tends to 0.and sketch a graph to support for answer.
MIT 18.01 Single Variable Calculus (OCW)
katieb
  • katieb
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anonymous
  • anonymous
\[lim_{x \rightarrow 0}(x+\frac{1}{x})=lim_{x \rightarrow 0}(\frac{x^2+1}{x})\] 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
anonymous
  • anonymous
as x tends to infinity the function is undefined as far as i know
anonymous
  • anonymous
use l'hopital rule

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anonymous
  • anonymous
|dw:1350219305254:dw|
anonymous
  • anonymous
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
anonymous
  • anonymous
graph is this and its correct
anonymous
  • anonymous
|dw:1350219633729:dw|
anonymous
  • anonymous
range of the grph is (-infinity , -2] union [2,infinity)
anonymous
  • anonymous
@adi171 there is no way i can use l 'HOpitals rule becoz the function gives 1/0 not 0/0 OR infnty/infnty.
anonymous
  • anonymous
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
anonymous
  • anonymous
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)

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