Thread: Find the limit - inverse trig

Find the limit:

lim x-> 0+ tan ^-1 (lnx) =?

I have no ide where to begin here? Can someone explain this in simple steps? Please, do not just give me another limit equation...
Thanks

lim x-> 0+ tan ^-1 (lnx) = tan^(-1)[lim(ln(x)] = tan^(-1)(-infinity)= -pi/2

Originally Posted by DBA

Find the limit:

lim x-> 0+ tan ^-1 (lnx) =?

I have no ide where to begin here? Can someone explain this in simple steps? Please, do not just give me another limit equation...
Thanks

I sure will give you another limit equation, son. If we let this becomes . Is the answer clear now?

Let's do something wacky and fun.

Note that





Use L'hopital on the inside:



The limit tends to -1 and we get



So, we get

Originally Posted by Calculus26

lim x-> 0+ tan ^-1 (lnx) = tan^(-1)[lim(ln(x)] = tan^(-1)(-infinity)= -pi/2

is lim x-> 0+ tan ^-1 (lnx) = tan^(-1)[limx-> 0+ (ln(x))] ?

Originally Posted by Drexel28

I sure will give you another limit equation, son. If we let this becomes . Is the answer clear now?

Hi, thanks.
but why lim z -> -infinity?

Originally Posted by galactus

Let's do something wacky and fun.

Note that





Use L'hopital on the inside:



The limit tends to -1 and we get



So, we get

Thanks, but we did not have L'hopital yet and with the number i I don't understand a thing. Sorry

yes in general

limf(g) = f(lim(g)) at a point if x = a if g is continuous at x = a and the limf exists at g(a)

I am sorry. Perhaps I shouldn't have done that. Just having fun. You will learn L'Hopital in time. It's a handy tool.

There are easier ways to go about it. Take note that