Example: Evaluate the limit

Is this expression defined on some reduced right neighborhood of a = 1? The only problems in it are the logarithm, which forces x > 1, and the sine in the denominator, which means that all integers are excluded from the domain. Thus the expression is defined also on the interval (1,2), which is a reduced right neighborhood of 1.

Consequently, we may try to evaluate this limit simply by substituting 1 into the given expression.

Conclusion: The limit diverges (to negative infinity, hence the limit exists).

Remark on one-sidedness in this example:

1. Concerning the logarithm: If x→1⁺, then also x > 1, therefore x − 1 > 0; that is, it is 0⁺.

2. Concerning the sine: If x→1⁺, then x is close to 1 and also x > 1. Consequently, πx is a number close to π satisfying πx > π. Therefore sin(πx) < 0 for such x and in the limit algebra it becomes 0^-.