# If f:R/{1,-1} -> R, f(x) = arctg[1/(x^2 - 1) calculate lim f(x) = ? when x->1 , x>1.

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lim f(X) = lim arctg (1/(1-x^2) when x-->1

Substitute with x=1

lim f(x) = arctg lim (1/0)

But we know that arctg = arcsin/arccos

==> arctg (1/0) = arc(sin(pi/2)/cos(pi/2)

= arctg(pi/2)

= pi/2

To find the lt arc(1/x^2-1) as x--> 1+

Solution:

We know that lt tan (x)-) asx-->(pi/2- ) is infinity. Therefore,

{Lt arc tan 1/(x^2-1) as x--> 1 } = lt arc tan ((pi/2)-) = pi/2.

lim arctg [1/(x^2 - 1)] = arctg lim [1/(x^2 - 1)] =

x -> 1 x -> 1

x > 1 x > 1

lim f(x) = arctg 1/(1-1) = arctg(1/+0) = arctg (inf) = **pi/2**