Problem: Evaluate (if it exists) the limit

Solution: This is a standard problem, we want to find a limit at infinity of an expression that exists on a neighborhood of infinity (e.g. for x > 0). Thus we start by substituting infinity into the expression.

How did we get those one-sided things? 1/∞ = 0 and if x goes to infinity, then x > 0, so also 1/x > 0. Thus we get 0⁺, that is, in the sine we have numbers that are close to zero and positive. Then also the sine of such numbers is positive and sin(0) = 0, so we get 0⁺ in the denominator.

Anyway, we have an indeterminate difference and the recommended method is to use algebra to put the two expressions together. Here the only reasonable way is to use the common denominator.

We need to find out the outcome of the indeterminate product in the numerator. There are two possible outcomes. We know that the denominator is 0⁺, so if the product in the numerator goes to a limit that is not 1 (even infinity is fine), then we can do the rest of the problem using the limit algebra. On the other hand, if the product goes to 1, then as a whole the problem is "zero over zero" and we have to do much more work.

So we now look at the product individually. The recommended approach to indeterminate products is to change them into a ratio, which will be also indeterminate and we can use l'Hospital's rule (or in general some other trick, but here no other way seems possible).

So the worst case scenario came true and we face an indeterminate ratio. By the way, note how we switch between 1/x (when we want to evaluate) and x⁻¹ (when we want to differentiate); such little tricks can make your life easier. Now it is time to return back to the ratio which is - as we just found out - indeterminate. No algebraic trick for cancelling seems possible, sine is not a part of intuitive reasoning, so it looks like another l'Hospital.

Similar problem, we have a ratio and in it we have an indeterminate product. The good news is that the denominator does not cause troubles and we can evaluate it separately. What can we say about the indeterminate product? It is a similar problem to the one we had before, just one extra power of x in it, so we can solve it exactly the same way. A smarter trick is to actually use the above calculation:

x²sin(1/x) = x⋅xsin(1/x)→∞⋅1 = ∞.

Thus the given problem transforms to another indeterminate difference.

Again, we need some way to transform this expression into a ratio. There is no obvious way like common denominator, what can we do? One possibility is to try the universal trick for changing products into ratios, but that would lead to dreadful things (as usual), if you are curious, look here.

What else can we do? There was a suggestion in the box indeterminate difference that can help here, namely the idea of factoring out something. Here we have a natural candidate: We factor out x. This will turn this difference into an indeterminate product.

By the way, we used the fact established above that xsin(1/x) goes to 1. This product has to be changed into a ratio and we have a natural candidate for "putting under", the x.

And so we ended with exactly the same expression that we started with except for the sign. What does it mean? We see that we reached a dead end, The approach that we adopted ends here. What can we do? One possibility is to use the factoring out right at the very beginning, unfortunately this is no big improvement.

So this does not help very much. How about another method from our arsenal? We do not have much luck: We can't cancel anything, no intuitive guessing suggest itself, nothing can be applied in those steps above. Are we beat?

Actually, no. The trick here is to start differently and solve a different problem. We note that at the root of the trouble is the chain rule, when we differentiate the composed sine and cosine. We have a trick up our sleeve that can help with this: substitution. If we set y = 1/x, then the sine will not be a composed function. How will it go?

How simple. As you can see, substitution can be a very powerful tool.

Remark: This substitution trick is still quite reasonable. There is another way to get the answer, a bit tricky and long but interesting nevertheless. In the first two attempts (turning the difference into a ratio and a product) the biggest problem was that after every l'Hospital, the expression did not get simpler. The culprit was the 1/x term, after every derivative we had lots of 1/x² around. One way to remove this problem is to use substitution as above. Another way is this: If we manage to change all appearances of x into 1/x then after differentiating, the terms 1/x² will appear everywhere and we can cancel. Turns out that this can be done. The first approach (with ratio) goes like this:

The second approach (with product) goes like this:

So it is possible to do this problem without substitution, but it is trickier and calculations nastier.

Remark: Note that the failed l'Hospital that returned back to the beginning can be used to get some conditional info. If the limit exists (and that was a big if at that moment), then we can call it L and the last failed step says L = −L. Thus if the limit exists, it can only be L = 0. The existence is a problem, but sometimes one can show konvergence using some theory.

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