# Show that $\lim_{n \to \infty} \frac{6n^3+2n^2-7}{(n+\sin(n^2))(n^2+1)}=6$.

Show that $$\lim_{n \rightarrow \infty} \frac{6n^3+2n^2-7}{(n+\sin(n^2))(n^2+1)} =6$$

I'm getting confused in the computation for $N$.

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Recall that if $-\infty < \displaystyle \lim_{n \to \infty} f(n) < \infty$, $-\infty < \displaystyle \lim_{n \to \infty} g(n) < \infty$ and $\lim_{n \to \infty} g(n) \neq 0$, then $$\displaystyle \lim_{n \to \infty} \dfrac{f(n)}{g(n)} = \dfrac{\displaystyle \lim_{n \to \infty} f(n)}{\displaystyle \lim_{n \to \infty} g(n)}$$ In your case, rewrite $$\dfrac{6n^3 + 2n^2 + 7}{(n+ \sin(n^2)) (n^2+1)}$$ as (by dividing the numerator and denominator by $n^3$) $$\dfrac{6 + 2/n + 7/n^3}{(1+ \sin(n^2)/n) (1+1/n^2)}$$

Now let $f(n) = 6 + 2/n + 7/n^3$ and $g(n) = (1+ \sin(n^2)/n) (1+1/n^2)$ and get the desired result.

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Should be some small caveat on the limit of $g(n)$ not being zero, yes? – rschwieb Sep 20 '12 at 16:01
@rschwieb Thanks. Updated. – user17762 Sep 20 '12 at 16:03

If you'd multiply the bottom out $(n+\sin(n^2))(n^2+1)$, you will have a term with the highest exponent $n^3$, then $\frac {6n^3}{n^3}$ =6

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HINT: consider the dominant terms of numerator and denominator and you're done.

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