Trig limits question

How would you do these with work

Lim xsin2x/1-cosx I think you would rationalize first....
x->0

and

lim sin(x-1)(x^2)/x^4-1
x->1

The first one should be like this:

Lim (x->0) xsin2x/(1-cosx)

Multiply by (1+cosx) / (1+cosx) so the the denominator turns into a square difference that equals sin²x. In addition use the following formula: sin2x = 2sinxcosx:

Lim (x->0) 2xsinxcosx(1+cosx)/sin²x = Lim (x->0) 2xcosx(1+cosx)/sinx

Now you know that Lim (x->0) x/sinx = Lim (x->0) sinx/x = 1 and you have

Lim (x->0) 2cosx(1+cosx) = 2*(1+1) = 4

The second one should be like this:

lim (x->1) sin(x-1)x²/(x⁴- 1) = lim (x->1) sin(x-1)x²/(x² - 1)(x² + 1) =
lim (x->1) sin(x-1)x²/(x - 1)(x + 1)(x² + 1)

Let x - 1 = t with t -> 0 and x = t + 1

lim (t->0) sin(t)(t + 1)²/t(t + 2)((t + 1)² + 1)

Use the above limit again and you have

lim (t->0) (t + 1)²/(t + 2)((t + 1)² + 1) = (0+1)²/(0+2)((0+1)² + 1) = 1/2*(1+1) = 1/4