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

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### 1 Answer

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.**