# Calculate the limit of the sequence $(\frac{n^2 + 2n}{3n^2 - 4})_{n\in\mathbb{N}}$ by using the algebra of limits

I know how to calculate the limit, however i haven't approached one of these that start with a negative number ($3n^2 -4$) in the denominator and also in which the denominator cannot be factorized in order to cancel out the numerator. Can someone please help with the first step of approaching this negative number.

• Divide numerator & denominator by $n^2$ – Empty Aug 25 '16 at 14:58
• You seem to consider it important that the denominator is negative for n= 1. Why is that? Do you understand that the first (any finite number) of terms is irrelevant to the limit of a sequence? – user247327 Aug 25 '16 at 15:00

My first step would be divide the numerator and denominator by $n^2$. $$\lim_{n \rightarrow \infty} \frac{n^2+2n}{3n^2-4}=\frac{\lim_{n \rightarrow \infty} (1+2/n)}{\lim_{n \rightarrow \infty}(3-4/n^2)}$$
$$\lim_{n\to\infty}(\dfrac{n^2 + 2n}{3n^2 - 4})=\lim_{n\to\infty}(\dfrac{1 + 2/n}{3 - 4/n^2})$$
Now limit of both numerator and denominator exists and limit of denominator is nonzero. Hence $$\lim_{n\to\infty}(\dfrac{n^2 + 2n}{3n^2 - 4})=\lim_{n\to\infty}(\dfrac{1 + 2/n}{3 - 4/n^2})=\dfrac{\lim_{n\to\infty}(1 + 2/n)}{\lim_{n\to\infty}(3 - 4/n^2})=1/3$$