lim_x->0 (x + (1/x) ) sin(x)(adsbygoogle = window.adsbygoogle || []).push({});

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)?

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# Another limit problem.

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