Question

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

Answer 1

`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`

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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?