lim  (sin 2x + 2) / 3x  Please show work.
x->0

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Request for Question Clarification by mathtalk-ga on 07 Jan 2004 08:27 PST

Hi, cousinit-ga:

Perhaps a few more parentheses, just to clarify what you are asking... is it:

lim  ( sin(2x) + 2 ) / (3x)
x->0

thanks, mathtalk-ga

Hi!!

In the case that the problem is:
lim  ( sin(2x) + 2 ) / (3x)
x->0

The solution is:
( sin(2x) + 2 ) is bounded between 1 and 3, because sin(2x) is bounded
between -1 and 1.

Then:

lim  1/(3x) =< lim  (sin(2x)+2)/(3x) =< lim  3/(3x)    (a)
x->0+          x->0+                      x->0+

         
and

lim  1/(3x) >= lim  (sin(2x)+2)/(3x) >= lim  3/(3x)    (b)
x->0-          x->0-                    x->0-


In the case (a) both extremes tend to +oo (positive infinitum); then
the right limit is +oo.
In the case (b) both extremes tend to -oo (negative infinitum); then
the left limit is -oo.
According to this the limit does not exist.

------------------------------------------------

If the problem is:
lim   sin(2x + 2) / (3x)
x->0


We have a similar situation because when x is close to zero we have that:
1/2 < sin(2x+2) < 1.

Then:
lim 1/2.(3x) =< lim   sin(2x+2)/(3x) =< lim 1/(3x)
x->0+           x->0+                   x->0+

and
lim 1/2.(3x) >= lim   sin(2x+2)/(3x) >= lim 1/(3x)
x->0-           x->0-                   x->0-


Proceeding like in the first case we will arrive to the same conclusion:
the limit does not exist.


I hope this helps you, if you find something unclear please ask for a
request of an answer clarification before rate this answer.

Best regards.
livioflores-ga

Very thorough.  I accidently entered the problem ambiguously (it could
be read as two different problems). The reseacher answered each
possibility and explained the steps clearly.  Thank you.