How do you find the limit of sin(x^2−4)/(x−2) sin(x24)x2 as x approaches 2?

2 Answers
Dec 14, 2016

Write it in a form that allows us to use lim_(thetararr0)sintheta/theta = 1

Explanation:

It looks like we want theta = x^2-4, so we'll multiply by (x+2)/(x+2), to get

((x+2)sin(x^2-4))/(x^2-4)

lim_(xrarr2)sin(x^2-4)/(x-2) = lim_(xrarr2)(x+2) * lim_(xrarr2)sin(x^2-4)/(x^2-4)

= (2+(2)) * (1) = 4

Here is the graph.

graph{sin(x^2-4)/(x-2) [-4.305, 8.185, -1.205, 5.04]}

Dec 14, 2016

lim_(x->2) sin(x^2-4)/(x-2) = 4

Explanation:

Substitute t=x-2

sin(x^2-4)/(x-2) = sin ((x+2)(x-2))/(x-2) = sin (t(t+4))/t=sin(t^2+4t)/t= frac (sint^2cos4t+cost^2sin4t) t = tcos4tsint^2/t^2 +4cost^2(sin4t)/(4t)

lim_(x->2) sin(x^2-4)/(x-2) = lim_(t->0) tcos4tsint^2/t^2 +4cost^2 frac (sin4t) (4t) = 0*1*1+4*1*1=4