What value of k makes the following function continuous?

Question

f(x)   =

 

sin(-6x)/(9x).   x is less than 0

 

3x-9k-6.  x is greater than or equal to 0

 

Help!?

Answer 1

Here is an approach without using L'Hopital's Rule:

 

We need lim_x→0^-  f(x) = lim_x→0⁺  f(x)

 

So, lim_x→0^-  [sin(-6x)/(9x)] = -9k - 6

 

     (1/9) lim_x→0^- [-sin(6x)/x] = -9k - 6  (the sine function is odd)

 

     (1/9)lim_x→0^-  [-6sin(6x/(6x)] = -9k - 6

 

     (-2/3)lim_x→0^-  [sin(6x/(6x)] = -9k - 6

 

     (-2/3)(1) = -9k - 6

 

     -9k - 6 = -2/3

 

            -9k = 16/3       So, k = -16/27

 

Answer 2

Use L'Hospital Rule on the trig expression.

 

-6cos(-6x) / 9 = -(2/3)cos(-6x)

 

 

Set this equal to the linear expression when x=0.

 

(-2/3) = -9k - 6

 

 

Solve for k.

 

Multiply both sides of the equation by 3.

 

-2 = -27k - 18

 

16 = -27k

 

-16/27 = k