Another limit problem.

[Dr-NiKoN]

lim_x->0 (x + (1/x) ) sin(x)

As x goes towards 0, wouldn't 1/x go towards inf?
Thus
(x + (1/x)) should go towards inf as x goes towards 0?

sin(x) will go towards 0 as x goes towards 0, since sin(0) is 0.

Wouldn't that leave inf * 0 = undefined (or 0)?

[arildno]

Note that your expression is the sum of two other expressions:
(x+\frac{1}{x})\sin(x)=x\sin(x)+\frac{\sin(x)}{x}
1. If it can be shown that each of these expressions (on the right-hand side) has a limit as x->0, what's then true about the limit of their sum (i.e, your original expression)?

[Dr-NiKoN]

lim_x->0 f(x) = x*sinx = 0
lim_x->0 g(x) = sinx/x = 1

lim_x->0 f(x) + g(x) = 0 + 1

lim_x->0 (x + 1/x))sinx = 1

Hm, that was a lot easier than I thought, thanks a lot :)

Edit: How do you do latex?
test:
\pi

nevermind :)

[modmans2ndcoming]

ah, just l'hopital the second part :-)