Wednesday, January 21, 2015

Evaluate the limit of the function y=ln(x^3+2x^2+3)/ln(x^4+x^2+2), x approaches to + infinite.

We'll verify first if by replacing x by accumulation
point, we'l get an indetermination.


y = ln infinite/ln
infinite = infinite/infinite


We'll factorize by x^3 at
numerator and by x^4 to denominator.


y = ln [x^3*(1 + 2/x +
3/x^3)]/ln [x^4*(1 + 1/x^2 + 2/x^4)]


We'll apply the
product rule of logarithms:


y = [ln x^3 + ln (1 + 2/x +
3/x^3)]/[ln x^4 + ln (1 + 1/x^2 + 2/x^4)]


We'll apply power
rule of logarithms:


y = [3*ln x + ln (1 + 2/x +
3/x^3)]/[4*ln x + ln (1 + 1/x^2 + 2/x^4)]


We'll factorize
both numerator and denominator by ln x:


y = ln x*[3+ ln (1
+ 2/x + 3/x^3)/ln x]/ln x*[4+ ln (1 + 1/x^2 + 2/x^4)/ln
x]


We'll simplify and we'll
get:


y = [3+ ln (1 + 2/x + 3/x^3)/ln x]/[4+ ln (1 + 1/x^2 +
2/x^4)/ln x]


We'll evaluate the
limit:


lim y = lim [3+ ln (1 + 2/x + 3/x^3)/ln x]/[4+ ln (1
+ 1/x^2 + 2/x^4)/ln x]


lim [3+ ln (1 + 2/x + 3/x^3)/ln
x]/[4+ ln (1 + 1/x^2 + 2/x^4)/ln x] =
3/4


After evaluation, the value of the limit
of the function y, if x approaches to + infinite, is: lim y =
3/4.

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