Find sin (x ^ 2-1) / (x-1) when x tends to the limit of 1,

Sin (x ^ 2-1) when x tends to be 1 equal to 0, it is an infinitesimal. Replace SiNx - > x with an equivalent infinitesimal, so sin (x ^ 2-1) - > (x ^ 2-1), (x ^ 2-1) / (x-1) = x + 1. When x tends to 1, x + 1 tends to 2, so the answer is 2
That's the limit
Of course, it's easy to use the lobita rule~

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