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

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Answer 1 · 2016-12-14T19:15:22

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

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]}

Answer 2 · 2016-12-14T19:50:37

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

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#