How do I find the following limit as x approaches π/2? (tanx)/(ln(x-π/2))?
I know I have to use l' hopital's rule at least twice, but I am doing something wrong somewhere and can't get the limit into a determinate form.
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y = tan(x) / (ln(x-π/2))
As x approaches π/2, tan(x) approaches ∞, ln(x-π/2) approaches -∞.
Apply l'Hôpital's rule:
Limit:
sec²(x) / (1/(x-π/2)) as x approaches π/2
= (x-π/2)sec²(x) as x approaches π/2
As x approaches π/2, (x-π/2) approaches 0.
As x approaches π/2, sec²(x) approaches ∞.
Apply l'Hôpital's rule again:
Limit:
2sec²(x)tan(x) as x approaches π/2
sec²(x) approaches ∞ as x approaches π/2
tan(x) approaches ±∞ as x approaches π/2
There is no (finite) limit.