limit of ln(sin x) as x approaches pi to the left
wouldn't lim ln(f(x)) be equal to ln Lim (f(x))?
To find the limit of ln(sin x) as x approaches pi from the left, we can begin by evaluating the limit of sin x as x approaches pi from the left.
The function sin x oscillates between -1 and 1 as x approaches pi from the left. Since the natural logarithm function ln(x) is undefined for x ≤ 0, we need to ensure that the argument of ln(sin x) is positive.
Since the range of sin x is [-1, 1], ln(sin x) is defined and positive if sin x is positive. Therefore, we need to find the limit of sin x as x approaches pi from the left and take the natural logarithm of that limit.
The limit of sin x as x approaches pi from the left is 1. Therefore, we have ln(sin x) = ln(1) = 0.
Thus, the limit of ln(sin x) as x approaches pi from the left is 0.