anonymous
  • anonymous
Help Please!!!(: Limit problems: -Suppose that f(x) denotes a function defined for all real numbers. Each of the statements below us true "sometimes". Give an example for each function to prove it holds a true and false function.
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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anonymous
  • anonymous
\[1.) \lim_{x \rightarrow 3} f(x)= f(3)\] \[2.) \lim_{x \rightarrow 0}\frac{ f(x) }{ x} = 1 \] then f(0)=0
DebbieG
  • DebbieG
Is the "then f(0)=0" part of #2?
anonymous
  • anonymous
yes

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DebbieG
  • DebbieG
For 1, the "true" part should be easy enough, right? Do you need help with that?
anonymous
  • anonymous
yes please!(:
DebbieG
  • DebbieG
You just need a function that's continuous at x=3. Take any polynomial. The simpler, the better. :)
DebbieG
  • DebbieG
Now, for it be FALSE, the easy way is just to make a piecewise function. Use something continuous (again, any polynomial would work) for \(x \neq3\), and then just define the function for x=3 so that it isn't = to f(3).
anonymous
  • anonymous
could you give me an example of a polynomial i could use?
DebbieG
  • DebbieG
Really, same with #2. You just need a continuous function such that f(0)=1 (a linear function would work fine, or something like x^2+1) Use that to define the function for all \(x\neq 0\). Then define f(0)=0.
DebbieG
  • DebbieG
Again, a linear would work.... or a quadratic..... You could even use absolute value. anything that's continuous at x=3 should work. Can you think of a function, and I'll let you know if it works?
anonymous
  • anonymous
could you help me with two more? It makes much more sense when you explain it (:
DebbieG
  • DebbieG
Did you get this one?
DebbieG
  • DebbieG
I'll help if I can, but you really should post each as a separate question, it just works better that way. I'll be on and off the site, I have to get my kids up for school soon. :)
DebbieG
  • DebbieG
You can always tag me in a problem... so that I see it.
anonymous
  • anonymous
yes I understood these ones(: and okay thank you!(:
DebbieG
  • DebbieG
you're welcome. :)

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