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Calculus/Limit Proof

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Question

Problem #111
I am working on my summer assignment for AP Calc BC, and I am stuck on how
to prove this.  It is question #111 in the picture attached.  In case you cannot
read it, it says: "Prove that if the limit of f(x) as x approaches c is 0 and the
absolute value of g(x) is less than or equal to M for a fixed number M and all x
does not equal c, then the limit as x approaches c of f(x)g(x) equals 0."

Answer
Let's prove it by the definition of limit . Reminder :
Lim  f(x) = 0 , means :
x->c
for every δ>0 exists ε>0 , so that for evey neighborhood of x
|x-c|<δ, we gain |f(x)|<ε . Where ε depend on δ (" ε(δ) ") .
We know that g(x) is bounded by M for all x, except x=c. We asume
that at x=c g(x) is defined. ("that means g(c)≠±∞"). Let's suppose
that g(c)=N. We define Q as : Q=Max{M,N}.
We can claim the following :
0 ≤ |f(x)g(x)| ≤ |Qf(x)| for all x in the motherhood |x-c|<δ .
We can also claim that -ε ≤ |f(x)g(x)| ≤ Q|f(x)| . That means for
every x such as |x-c|<ε , we gain : -ε ≤ |f(x)g(x)| ≤ Qε .
Let's define ε'=Qε , therefore we can state that :
-ε' ≤ |f(x)g(x)| ≤ ε' , which is the same as |f(x)g(x)|<ε' for
every |x-c|<δ. Hence we proved that
Lim  f(x)g(x) = 0   
x->c

Alon.

Alon Mandes

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