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

Answer 1

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

#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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Answer 2

To evaluate the limit lim(x+sinx)dx as x approaches 0, we can use the limit properties and trigonometric identities. By substituting 0 into the expression, we get 0 + sin(0) which simplifies to 0 + 0 = 0. Therefore, the limit of (x+sinx)dx as x approaches 0 is equal to 0.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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