lim arcsec (x)/√(x-1)
x-1

I know from using L'Hospital's Rule that this is √2, but how do I do this w/o that rule or using derivatives?

Question

lim arcsec (x)/√(x-1)
x->1

I know from using L'Hospital's Rule that this is √2, but how do I do this w/o that rule or using derivatives?

rogue · Answer

$$\lim_{x \rightarrow 1} (\cos^{-1} (1/x)) / \sqrt{x - 1})$$

rogue · Answer

There is a way to do this using many substitutions but it often gets complicated.

rogue · Answer

$$\mu = \sqrt{x - 1}, \mu ^{2} = x - 1, x = \mu ^{2} + 1$$

anonymous · Answer

I see what you're doing, but even that seems a little complex for a beginning calc student. I was helping a kid with his homework last night and that question was there. For the life of me I can't figure out how they were supposed to do it

rogue · Answer

$$\lim_{\mu \rightarrow 0} (Arcsec (\mu ^{2} + 1))/\mu)$$

rogue · Answer

Yeah, for introductory Calc its very complicated, the teacher may have made a mistake or something. My teacher gave a do now with a question like that this year, but it was with regular trig functions, not inverse trig, so it was solvable via substitution. This question requires some esoteric identities if one wanted to solve this without L'hospital's.

anonymous · Answer

Right, at this stage they're really limited in what they can do with inverse trig functions. Thanks