# Evaluate the limit of the product f(x)*g(x) if f(x)=x^2/1+x^2, g(x)=arctanx x-->+infinite

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f(x)= x^2/ (1+x^2)

g(x) = artanx

We need to calculate:

lim f(x)*g(x) x--> +inf

lim (x)*g(x)= lim [x^2/(1+x^2)*arctanx]

= lim (x^2/(1+x^2) * lim arctanx

Divide and multiply by x^2:

==> lim (1/1/x^2 + 1) * arctan limx

when x -->

==> lim f(x)*g(x) = 1 * arctan (inf)

= 1*pi.2 = pi/2

==> lim f(x) *g(x) = pi/2

To evaluate the limit, we'll apply the rule that the limit of the product is the product of limits:

lim f(x)*g(x)=lim f(x)*lim g(x)

Let's calculate lim f(x):

lim x^2/(1+x^2)=lim x^2/ lim [x^2(1/x^2 + 1)]

We'll cancel the factor x^2:

lim 1/ lim(1/x^2 + 1)]=1/(lim(1/x^2 + lim 1)= 1/(0+1) = 1

Let's calculate lim g(x):

lim arctg x= arctg limx =arctg infinity= pi/2

**lim f(x)*lim g(x) = 1*pi/2 = pi/2**

f(x) = x^2/(1+x^2) and g(x) - arctanx.

To find lt f(x)*g(x) as x --> inf.

Solution:

f(x)*g(x) = x^2arctanx/(1+x^2) = {(1+x^2)(arctanx)- arctanx}/(1+x^2) = arctanx - arctanx/(1+x^2).

Therefore

Lt x^2 arc tanx/(1+x^2) = lt arctanx - ltarctanx/(1+x^2)

= pi/2 - (pi/2)/(1+inf)

= pi/2 -0

=pi/2.