Find the value of K. Use of l Hospital's rule and expansion is not allowed.Let...

Jovany Clayton

Jovany Clayton

Answered

2022-07-12

Find the value of K. Use of l Hospital's rule and expansion is not allowed.
Let f ( x ) = log c o s 3 x ( cos 2 i x ) if x 0 and f ( 0 ) = k where ( i= iota) is continuous at x = 0, then find the value of K. Use of l Hospital's rule and expansion is not allowed.

Answer & Explanation

1s1zubv

1s1zubv

Expert

2022-07-13Added 17 answers

(Preamble: Apart from addition and multiplication the only genuine two-ary function allowed in print is the general power function:
(1) ( x , y ) x y := exp ( y log x ) ( x > 0 , y C )   .
The function ( x , y ) log x y where both x and y are variables is not of this kind. When such a thing should really come up in practice one would have to rewrite it in such a way that only functions of one variable and the "allowed" operations appear. It is true that we sometimes write log a y. But the a > 1 here is not a variable; it is a constant scaling parameter chosen in a most suitable way for the problem at hand.)
So after a lot of mind bending I'm interpreting your problem as follows: For 0 < | x | 1 one has 0 < cos ( 3 x ) < 1. On the other hand
cos ( 2 i x ) = e 2 x + e 2 x 2 = cosh ( 2 x ) ( x R )   .
Therefore the equation
( cos ( 3 x ) ) y = cos ( 2 i x ) = cosh ( 2 x )
has a real solution y for such x; and according to ( 1 ) this y is given by
(2) y = log ( cosh ( 2 x ) ) log ( cos ( 3 x ) ) =: f ( x ) ( 0 < | x | 1 )   .
The right hand side of ( 2 ) is the correct expression of the function f your teacher had in mind.
Now it comes to computing lim x 0 f ( x ). Since both de l'Hôpital's rule and expansion are forbidden I have to stop here. It is unclear to me how this limit can be computed without using the analytical properties of the involved functions somehow.

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