How do you find the limit of #(x cos (1/x))# as x approaches #oo#? Calculus Limits Determining Limits Algebraically 1 Answer Jim H Jun 1, 2016 See below. Explanation: As #xrarroo#, we have #cos(1/x)rarr1#. And for #x# sufficiently great (#x > 2/pi#) , we have #0 < cos(1/x)#, so #co(1/x)# is positive and approaching #0#. As #x# increases without bound, #xcos(1/x)# approaches #x# which is (of course) increasing without bound. #lim_(xrarroo)xcos(1/x) = oo# Answer link Related questions How do you find the limit #lim_(x->5)(x^2-6x+5)/(x^2-25)# ? How do you find the limit #lim_(x->3^+)|3-x|/(x^2-2x-3)# ? How do you find the limit #lim_(x->4)(x^3-64)/(x^2-8x+16)# ? How do you find the limit #lim_(x->2)(x^2+x-6)/(x-2)# ? How do you find the limit #lim_(x->-4)(x^2+5x+4)/(x^2+3x-4)# ? How do you find the limit #lim_(t->-3)(t^2-9)/(2t^2+7t+3)# ? How do you find the limit #lim_(h->0)((4+h)^2-16)/h# ? How do you find the limit #lim_(h->0)((2+h)^3-8)/h# ? How do you find the limit #lim_(x->9)(9-x)/(3-sqrt(x))# ? How do you find the limit #lim_(h->0)(sqrt(1+h)-1)/h# ? See all questions in Determining Limits Algebraically Impact of this question 1239 views around the world You can reuse this answer Creative Commons License