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anonymous
 5 years ago
Prove the statement using the epsilon, sigma definition of limit.
lim (73x)=5
x>4
anonymous
 5 years ago
Prove the statement using the epsilon, sigma definition of limit. lim (73x)=5 x>4

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I am struggling with these. :( Would be nice if some1 could show it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you should start with something like this: for all epsilon (I will use E) >0 there exist a delta (D) such that D>x4 implies E>f(x)(5) Or something like this

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but what does that equation mean and how do you solve it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0good question :D I am looking at my notes now and trying to figure it out but not working well

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I am doing real analysis, are u too?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'll be back in 10 min and give it another go

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So the definition is \[given any \epsilon >0 there \exists \delta >0 such that 0<\left xa \right<\delta \implies \left \left f(x)L \right \right<\epsilon\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok so all you do is plug in the 4 for x which gives the answer as 712=5, 5=5 and that is it. Source a friend who got an a in calc 3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes that is fine, but it does not use epsilon or delta...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I guess here you need to use this definition but I'm not sure how. :(
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