How do you evaluate the limit #lim(2x-cosx)dx# as #x->0#?

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Answer 1 · 2018-03-01T04:06:44

#-1#

Let's split up our limit, as #lim_(x->a)(f(x)+-g(x))=lim_(x->a)f(x)+-lim_(x->a)g(x)#

#lim_(x->0)(2x-cos(x))=lim_(x->0)2x-lim_(x->0)cos(x)#

Let's plug #0# into each term:

#lim_(x->0)2x-lim_(x->0)cos(x)=2(0)-cos(0)=0-cos(0)#

Note that we drop the limit notation as soon as we start plugging in the value we're approaching.

#cos(0)=-1#, so we can say:

#lim_(x->0)(2x-cos(x))=0-1=-1#