Use a table of values to estimate the value of the limit

Question

Use a table of values to estimate the value of the limit (correct to three decimal places). If you have a graphing device, use it to confirm your result graphically.

Answer 1

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

-> sin(2x + x) / [sin(2x)/cos(2x)]

= cos(2x)[sin(2x)cos(x) + sin(x)cos(2x)] / sin(2x)

= cos(2x)[ cos(x) + ( sin(x)cos(2x) / 2sin(x)cos(x) ) ]

= cos(2x)[ cos(x) + ( cos(2x) / 2cos(x) ) ]

= cos(2x)[ cos(x) + (2cos²(x) - 1) / 2cos(x) ]

= cos(2x)[ cos(x) + cos(x) - (1 / 2cos(x)) ]

= [2cos²(x) - 1][ 2cos(x) - 1 / 2cos(x) ]

= [4cos³(x) - cos(x) - 2cos(x) + 1 / 2cos(x) ]

= [4cos³(x) - 3cos(x) + 1 / 2cos(x) ]

lim(x->0) sin(3x)/tan(2x) = lim(x->0) [4cos³(x) - 3cos(x) + 1 / 2cos(x) ] = 4 - 3 + 1/2

==> lim(x->0) sin(3x)/tan(2x) = 3/2

Table:

(We are near x=0, so with high frequencies 2x, and 3x, so the period is fairly small… we will not need to go far to find “Zeroes" )

Question... When is:

4cos³(x) - 3cos(x) + 1 / 2cos(x) = 0 ?

4cos³(x) - 3cos(x) = -1 / 2cos(x) 8cos⁴(x) - 6cos²(x) + 1 = 0

# let u = cos²(x) 8u² - 6u + 1 = 0

==> u = (1/16)[ 6 ± √(36 - 32) ] = (1/16)(6 ± 2) = 1/2, 1/4

u = cos²(x) = 1/2, 1/4 ==> cos(x) = 1/√2 , 1/2 ==> x = ± π/4, ± π/3 (i.e. 45º, 60º) ==> x = ± .7854 , ± 1.0467

Radians

Table:

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

-π/3 0

-π/4 0

-0.1 1.458

-0.05 1.489

0 3/2

0.05 1.489

0.1 1.458

π/4 0

π/3 0

Yeah, I thought you might enjoy that ! I bet I had you worried with all the 0 before convergence, then suddenly 3/2 ! So, I added ± 0.1 and ± 0.05 (Radians) around 0.

(Can’t paste Desmos graph below… )

Enter y = sin(3x)/tan(2x), then zoom in...

Answer 2