Calculate the limit of these sequences;
a) $$a_n=\dfrac{1-3n}{n^2+2}$$
b) $$a_n=\dfrac{4n^2+5n+6}{6n+8}$$
c) $$a_n=\dfrac{7n^3-11}{-2n^3-76}$$
d) $$a_n=\dfrac{5^n}{(-3)^n}$$
Development:
a) As the grade of the polynomial of the numerator is smaller than that of the denominator the limit is $$0$$.
b) If the grade of the polynomial of the numerator is greater than that of the denominator then, in this case, the sequence tends to infinity. To calculate the sign we look at the sign of the quotient of the coefficients of the largest grade of the two polynomials. This quotient corresponds to $$\dfrac{4}{6}$$ which is positive. Therefore the limit of the sequence is $$+\infty$$.
c) As the grades of the polynomial of the numerator and that of the denominator are equal, the limit corresponds to the quotient of the coefficients of largest grade of the two polynomials. In this case the coefficient of the largest grade of the numerator is $$7$$ and that of the denominator is $$-2$$. This way the limit of the sequence is $$-\dfrac{7}{2}$$.
d) The sequence corresponds to the sequence $$a_n=b^n$$ with $$b=-\dfrac{5}{3}$$. As $$-\dfrac{5}{3} \leq -1$$ and the sequence has no limit.
Solution:
a) The limit of the sequence is $$0$$.
b) The sequence tends to $$+\infty$$.
c) The limit of the sequence is $$-\dfrac{7}{2}$$.
d) The sequence has no limit.