2: You should get

Note that the moment there appeared ln(0), it was necessary to further inquire about the type of 0, going back to where it all started. Since 1/n > 0 for positive n, there is in fact sin(0⁺), and sine of small positive numbers is again positive, so sin(0⁺) = 0⁺.

Anyway, the limit expression you face now is an indeterminate product, so apply the standard procedure. Change the product into a ratio by putting something "under", probably the sine, since the logarithm would be more complicated; another reason is that leaving the sine on top would not improve it (derivative changes it into a cosine that behaves similarly), while leaving logarithm up will allow the derivative to get rid of it.

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