Find the limit as n goes to infinity of (3n² - 4n + 9)/(n² + 32n + 48) - the square root of (n² + n + 18).

Step-by-step solution

1. 1

   Split the expression and study each part separately. The two pieces behave completely differently, so there is no indeterminate form to resolve here - the trick is recognising that early rather than reaching for L'Hopital or a conjugate trick.

2. 2

   Find the limit of the rational term. Divide numerator and denominator by the highest power $n^2$: $\frac{3n^2-4n+9}{n^2+32n+48} = \frac{3-\tfrac4n+\tfrac{9}{n^2}}{1+\tfrac{32}{n}+\tfrac{48}{n^2}} \longrightarrow \frac{3}{1} = 3.$ Equal degrees top and bottom means the limit is the ratio of leading coefficients.

3. 3

   Find the behaviour of the square root. Factor $n^2$ out from under the radical: $\sqrt{n^2+n+18} = n\sqrt{1+\tfrac1n+\tfrac{18}{n^2}} \sim n \cdot 1 = n \to +\infty,$ using $|n|=n$ since $n\to+\infty$. More precisely it is $n+\tfrac12+O(1/n)$.

4. 4

   Combine the two behaviours. The expression is (something tending to $3$) minus (something tending to $+\infty$). A bounded quantity minus an unbounded one is unbounded below: $3-(+\infty) = -\infty.$ This is a determinate form, not an indeterminate one.

5. 5

   Confirm numerically. At $n=10^6$ the rational term is $2.99990$ and the root is $1{,}000{,}000.5000$, so the difference is about $-999{,}997.5$ - already tracking $-n$ as predicted.

6. 6

   State the conclusion. $\lim_{n\to\infty}\left(\frac{3n^2-4n+9}{n^2+32n+48}-\sqrt{n^2+n+18}\right) = -\infty,$ and asymptotically the expression behaves like $-n+2.5$.