Question: "Evaluate limx→0+arccos⁡(x)−π2xln⁡(x)"

Rowan Callahan

Answered question

2022-04-04

Evaluate

lim_{x \to 0^{+}} \frac{\arccos (x) - \frac{π}{2}}{x \ln (x)}

Kingston Lowery

Beginner2022-04-05Added 10 answers

lim_{x \to 0^{+}} \frac{\arccos (x) - \frac{π}{2}}{x \ln x} = lim_{x \to 0^{+}} \frac{\arccos (x) - \arccos (0)}{x - 0} \cdot \frac{1}{\ln x} = - \frac{1}{\sqrt{1 - 0^{2}}} \cdot 0 = 0

It is valid, because if we know that

lim_{x \to 0^{+}} f (x)

and

lim_{x \to 0^{+}} g (x)

exist, then we have that

lim_{x \to 0^{+}} f (x) g (x)

exists and

lim_{x \to 0^{+}} f (x) g (x) = lim_{x \to 0^{+}} f (x) lim_{x \to 0^{+}} g (x)

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