How do you find the limit of #sin((x-1)/(2+x^2))# as x approaches #oo#?

Question

1303 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-20T18:55:42

Factorise the maximum power of #x# and cancel the common factors of the nominator and denumerator. Answer is:

#lim_(x->oo)sin((x-1)/(2+x^2))=0#

#lim_(x->oo)sin((x-1)/(2+x^2))#

#lim_(x->oo)sin((1*x-1*x/x)/(2*x^2/x^2+1*x^2))#

#lim_(x->oo)sin((x*(1-1/x))/(x^2*(2/x^2+1)))#

#lim_(x->oo)sin((cancel(x)(1-1/x))/(x^cancel(2)(2/x^2+1)))#

#lim_(x->oo)sin((1-1/x)/(x(2/x^2+1)))#

Now you can finally take the limit, noting that #1/oo=0#:

#sin((1-0)/(oo*(0+1)))#

#sin(1/oo)#

#sin0#

#0#