Calculate the limit of $\sqrt{n^2-3n}-n$ as $n \rightarrow \infty $

Question

Calculate the limit of $\sqrt{n^2-3n}-n$ as $n \rightarrow \infty $.

I'm having a hard time trying to simplify this expression in order to be able to get the limit of it.

I have $\sqrt{n^2-3n}-n=\sqrt{n^2(1-\frac{3}{n})}-n=n\sqrt{1-\frac{3}{n}}-n=n(\sqrt{1-\frac{3}{n}}-1)$

Don't think that this gives me anything useful, can anybody give me some hints.

Answer 1

Hint: multiply and divide by the ‟conjugate” expression $\sqrt{n^2-3n}+n$.

Answer 2

You can do as recommended above, or you can try this:

Pull the $n$ out of the square root to give $$(n^2 -2n)^{1/2} -n = n((1-2/n)^{1/2}-1) \\ =n[(1 + {1 \over 2}{-2 \over n} - {1 \over 4}({-2 \over n})^2 + \ ...) -1] = -1 - 1/n + \ ... $$ and thus when you take the limit as $n \rightarrow \infty$, you get the limit of $-1$.

Hope this helps - if it does, please remember to upvote and/or accept my answer! :)

Answer 3

Hint

You found

$$n(\sqrt{1-\frac{3}{n}}-1)$$ now use the Taylor approximation $(1+x)^\alpha\sim_01+\alpha x$ so $$n(\sqrt{1-\frac{3}{n}}-1)\sim_\infty n\times \left(-\frac12\frac 3n\right)=-\frac32$$

Answer 4

Another way: Write what you got as $\displaystyle \lim_{n\to\infty}\frac{\sqrt{1-\frac{3}{n}}-1}{\frac{1}{n}}$ and use L'Hopital's rule.

Answer 5

Hint

Another way : write $$\sqrt{n^2-2n}-n=n \sqrt{1-\frac{2}{n}}-n=n(\sqrt{1-\frac{2}{n}}-1)$$ and remember that, when $x$ is small compared to $1$, $$\sqrt{1+x}\simeq 1+\frac{x}{2}$$ So, replace in the last expression $x$ by $-\frac{2}{n}$, expand and multiply.

Answer 6

Try completing the square to give $$\sqrt {n^2-3n}-n=\sqrt {\left(n-\frac32\right)^2-\frac 94}-n$$ which you should be able to show goes to $$n-\frac 32-n=-\frac 32$$