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anonymous
 5 years ago
fine the limit if it exists:
lim of x>0 [[1/(3+x)]  (1/3)]/x
anonymous
 5 years ago
fine the limit if it exists: lim of x>0 [[1/(3+x)]  (1/3)]/x

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(1/9) is what I got taking the derivative to the top and bottom after some algebra..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it is 1/9 http://www.wolframalpha.com/input/?i=lim+of+x%3E0+ [[1%2F%283%2Bx%29]++%281%2F3%29]%2Fx And click "show steps"

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this is the derivative of \[f(x)=\frac{1}{x}\] at x = 3 but we can do it without the derivative, just with some algebra

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hhmmmm. the link gets broken. you'll need to copy/paste it .

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{3+x}\frac{1}{3}=\frac{3(3+x)}{x(3+x)}\] \[=\frac{x}{3(3+x)}\] divide by x get \[\frac{1}{3(3+x)}\] replace x by 0 get \[\frac{1}{9}\]
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