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
Follow Math Help Forum on Facebook and Google+
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
View Tag Cloud