Limit with sin (without hopital)

Question

limx-->2 ( sin(2x-4)/(x-2)

Answer 1

Note that 2x-4 = 2(x-2).  Then

 

sin(2x-4) / (x-2) = sin[2(x-2)] / (x-2)

 

Multiply by 2/2 and get

 

sin(2x-4)] / (x-2) = 2sin[2(x-2)] / [2(x-2)]

 

Let w = 2(x-2)

 

Then sin(2x-4) / (x-2) = 2 sin(w) / w

 

Then limit as x→2 of sin(2x-4) / (x-2) = limit as w→0 [2sin(w) / w]

 

Are you familiar with limit as w→0 of sin(w) / w = 1?

 

Use that to get a final limit of 2 for the initial expression.

Answer 2

lim_x→2[sin(2x-4)/(x-2)]

 

= lim_x→2[2sin(2x-4))/(2x-4)]

 

=  2lim_x→2[sin(2x-4)/(2x-4)]

 

       Let θ = 2x-4.  Then, as x→2, θ=2(x-2)→0

 

So, the given limit is equivalent to 2lim_θ→0[sinθ/θ]

 

                                             =  2(1) = 2

 

Answer 3

I would use u = x-2

 

lim u-->0 (sin 2u)/u

 

= lim u --> 0 (2sin u cos u)/u

= lim u --> 0 2 (sin u)/u * cos u

= 2*1*1 = 2