does anybody know what the limit would be for: lim(x->infinite) of [(pi/2) - arctanx]^(1/(lnx)) ?

Answer 1

the limit is equal to 1

Answer 2

First, in case you don't know it already, let's find the derivative of arctan x: set y = arctan x, so x = tan y. Then 1 = sec^2 y dy/dx, so dy/dx = 1/sec^2 y = 1 / (tan^2 y + 1) = 1 / (x^2 + 1).

Now, for f = [pi/2 - arctan x]^(1/ln x), look at
g = ln f = ln (pi/2 - arctan x) / ln x. Note that the numerator goes to -inf and the denominator to +inf as x goes to infinity. So apply L'Hopital's Rule to get
lim (x->inf) of [(1/(pi/2 - arctan x)). (1/(x^2+1))] / (1/x)
= lim (x->inf) of -x / [(x^2+1) . (pi/2 - arctan x)]
= lim (x->inf) of (-1/x) / (pi/2 - arctan x)
This is in 0/0 form, so we apply L'Hopital's rule again:
= lim (x->inf) of (1/x^2) / (-1/(x^2+1))
= lim (x->inf) of -(x^2+1)/x^2
= -1. Since this is a real number our previous two applications of L'Hopital's Rule were justified.

Now remember, this is the natural log of the original function. So if ln f -> -1, f -> e^(-1) = 1/e ~ 0.367879.

Answer 3

The exact answer is 1/e where e is the base of the natural logarithm which has an approximate value of 2.7182818.... if you are to get the value of 1/e that's approximately 0.367...

Answer 4

the limit is equal 1.

Answer 5

Y=(pi/2-atan(x))^(1/lnx), lny = ln(pi/2-atan(x))/lnx;
(1)Now lim of (pi/2-atan(x)) = pi, while lim of lnx = +inf; thus lim of lny = 0, hence lim of y = 1;
(2) x=cot(pi/2-atan(x)), z=acot(x), x=cot(z) z->pi/2
Lim of lny = ln(z)/ln(cot(z)) = (l’hopital) = -(cot(z)/z)*(sinz)^2 = 0 lim of y =1;
Both = 1

Answer 6

.36788