12: After substituting in the limit point you should get the type 0⋅∞⋅∞. This is an indeterminate product and no algebraic trick seems possible, so the standard approach should be used: Change the product into a ratio. However, there are three terms now. What to do?

There are essentially two possibilities. One is to take the whole infinity part "xln(x)" and "put it under". The advantage is that the resulting ratio will be for sure of the type "zero over zero", so the l'Hospital rule could be used; that is, there is no catch waiting (at least as far as we can see now). The disadvantage is that the expression that will be created in the denominator will be far from nice, and taking derivative will only make it worse.

What is the alternative? Take one "infinity" away, that is, split the limit into two. For instance, take the logarithm out, the limit of the logarithm is infinity. The second limit will be the arctangent part multiplied by x, which is still an indeterminate product, but now much simpler. Indeed, moving x "under" will create x⁻¹ there, which is easy to differentiate. What is the disadvantage of this approach? After evaluating the two limits, the answers would have to be put together. The first limit is infinity, so if the second comes out as infinity or some number not equal to zero, the limit algebra could be used to put the results together and all is well. However, if the second limit happens to be zero, the two results would make together an indeterminate product, so all the work would be in vain and one would have to try something else, for instance the first procedure outlined above.

Choose whichever way you feel like. The author is a chicken and decided to play it safe, the hints will follow the less risky (albeit perhaps longer) first approach. We will briefly outline the shorter but risky way as a note in the next hint. Try it, try also to pull out the x, it's a good practice.

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