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

1 Answer
Alan P.
Nov 16, 2017

#lim_(xrarr0) (x+sin(x))dx=0#

Explanation:

#lim_(xrarr0) (x+sin(x))dx#

#color(white)("XXX")=color(blue)(lim_(xrarr0)x * dx) +color(magenta)(lim_(xrarr0)sin(x) * dx)#

#color(white)("XXX")=color(blue)( 0 * dx)+color(magenta)(0 * dx)#

#color(white)("XXX")=0#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Is it possible that the original question should have been:
#color(white)("XXX")lim_(xrarr0) (d(x+sin(x)))/(dx)#
or something else along those lines?

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