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May 29, 2015

Homework Help: Calculus

Posted by Frederique on Sunday, November 19, 2006 at 9:36am.

Note that

pi
lim arctan(x ) = ----
x -> +oo 2


Now evaluate


/ pi \
lim |arctan(x ) - -----| x
x -> +oo \ 2 /

I'm not exactly sure how to attempt it. I have tried h'opital's rule but I don't believe you can use it here. Any help will be greatly appreciated!

Sorry the question came out weirder than i had originally posted it

it is the lim as x approaches positive infinity of (arctanx - pi/2)x



Use that:

arctan(x) = pi/2 - arctan(1/x)

If you take a right triangle then you can easily see where this relation comes from. If x is the ratio between two right sides then 1/x is the inverse of that ratio, so arctan(1/x) will yield the other angle which is pi minus arctan(x).

If x approaches infinity, 1/x approaches zero, so you can use the series expansion of the arctangent function around x = 0:

arctan(x) = x - x^3/3 + x^5/5 -...
for x in a neighborhood of zero --->

arctan(1/x) = x^(-1) - x^(-3)/3 + x^(-5)/5 -..

for x -->infinity


Therefore for large x:

arctan(x) = pi/2 - arctan(1/x)=

pi/2 - 1/x + 1/(3 x^3) - ...

And you can now read-off the desired limit :)

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