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Mathematics 7 Online
OpenStudy (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.

OpenStudy (anonymous):

\[1.) \lim_{x \rightarrow 3} f(x)= f(3)\] \[2.) \lim_{x \rightarrow 0}\frac{ f(x) }{ x} = 1 \] then f(0)=0

OpenStudy (debbieg):

Is the "then f(0)=0" part of #2?

OpenStudy (anonymous):

yes

OpenStudy (debbieg):

For 1, the "true" part should be easy enough, right? Do you need help with that?

OpenStudy (anonymous):

yes please!(:

OpenStudy (debbieg):

You just need a function that's continuous at x=3. Take any polynomial. The simpler, the better. :)

OpenStudy (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).

OpenStudy (anonymous):

could you give me an example of a polynomial i could use?

OpenStudy (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.

OpenStudy (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?

OpenStudy (anonymous):

could you help me with two more? It makes much more sense when you explain it (:

OpenStudy (debbieg):

Did you get this one?

OpenStudy (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. :)

OpenStudy (debbieg):

You can always tag me in a problem... so that I see it.

OpenStudy (anonymous):

yes I understood these ones(: and okay thank you!(:

OpenStudy (debbieg):

you're welcome. :)

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