SOLUTION: limit (((ln 2))^(x))/(cos ((pi)/4 sin (3 x))) as x \[LongRightArrow] - \[Infinity] = 0 ( T or F )

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Question 1208976: limit (((ln 2))^(x))/(cos ((pi)/4 sin (3 x))) as x \[LongRightArrow] - \[Infinity] = 0 ( T or F )
Answer by ikleyn(52767)   (Show Source): You can put this solution on YOUR website!
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limit (((ln 2))^(x))/(cos ((pi)/4 sin (3 x))) as x \[LongRightArrow] - \[Infinity] = 0 ( T or F )
~~~~~~~~~~~~~~~~~~~~~~~~~~~

I placed (copy-pasted) the given expression into the Math editor, and this Math editor interpreted the expression this way

    .


It is what I will use in my following analysis.


             Step 1.


Consider a sequence of real numbers  , , , . . . , , . . . which approaches to    from the left.

Then the sequence of real numbers  , , , . . . , , . . . approaches to    from the left.

Then the sequence of real numbers  , , , . . . ,   approaches to   = 1/2  from the left.

Then the sequence of real numbers  , , , . . . , , . . .  

         approaches to   =  =   from the left.


Then the sequence of numbers    approaches to  0 (zero)  from the right.


Thus for this sequence of numbers  ,  the numerator approaches to (ln(2))^(pi/18), which is some constant,

while the denominator    approaches to zero.


Hence, as the argument x approaches to    from the left,  
the values of the given expression go to positive infinity.


After that, as the argument x becomes greater than , the expression takes finite values, again.


             Step 2.


The function    in the denominator is periodical with the period .


It means that the behavior of the given expression, which we detected in part (a), repeats inside each period [,],
for all positive or negative integer numbers "n".


In other words, inside each such period, there is a converging sequence of real numbers, 
for which our expression goes to positive infinity, and after that the expression takes finite values, again.


             Step 3.


It means that as x goes to plus or minus infinity, the given expression HAS NO limit.


ANSWER.  Of the two possible options, T or F, only F is valid.

Solved.


========================


It can be solved in other way, too.

In my solution above, I considered a sequence of numbers  ,  converging to  pi/6  from the left,
and shoved that for such sequence  of arguments the whole function has positive values (going to + infinity).


Similarly, if we consider another sequence of numbers  , converging to pi/6 from the right,
we will get the sequence of the values of the whole function, which all are negative and go to - infinity.



Such a behavior repeats at each interval [,], for all positive or negative integer numbers "n".


But the function with such a behavior can not have limits as x goes to +infinity or -infinity.



So, the given function has no limits as x goes to +infinity or -infinity.



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