Okay I looked in my book and figured out some stuff. What you need to do is take the limit of the function as x approaches 0. That basically just means plug in 0 for x. You end up with 0/0, which is indeterminite. But you can apply L'Hopital's Rule to this because the indeterminite form is either 0/0 or infinity/infinity.
L'Hopitals Rule says that the limit as x approaches some number c of the numerator divided by the denominator is equal to the limit as x approaches c of the derivative of the numerator divided by the derivative of the denominator.
This is it:
\(\lim_{x \to c}\frac{f(x)}{g(x)}=\lim_{x \to c}\frac{f'(x)}{g'(x)}\)
Soooo in your case
f(x)=x2 + xsinx
g(x) = x2
c=0
then you need to find f'(x) and g'(x).
f'(x)=2x+sinx+xcosx
g'(x)= 2x
then just plug it in plug it in
\(\lim_{x \to 0}\frac{2x+sinx+xcosx}{2x}\)
If you apply L'Hopitals Rule again you get
\(\lim_{x \to 0} \frac{2+cosx-xsinx+cosx}{2}\)
then plug in 0 for x and you get
\(\frac{2+1-0+1}{2}\)
which is 4/2
which is.... dun dun dun..... 2!!!! :)))
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