Solving Lim [(xsin (π / x) + (π / x) SiNx)] I use the Equivalent Infinitesimal Substitution, the result of calculation is 2 π, but the answer is π. What's the matter? What's wrong? Can you list the calculation process in detail X→∞

The equivalent infinitesimal has a premise, that is, the substitution is infinitesimal. It is estimated that you did not pay attention to this condition, and the second term can be calculated directly. A bounded function multiplied by 0 results in 0, so it should be Lim [(xsin (π / x) + (π / x) SiNx)] = Lim [sin (π / x) / (1 / x)] + 0 = lim

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