Evaluating Limits - big fraction

Question

is anyone able to help me evaluate this limit :). What I have done so far is expand out the fraction by multiplying the top and bottom of the fraction by the conjugate of the numerator. But, this produced a very large fraction which doesn't seem like it would help. Thankyou

$$\lim_{n\to \infty} \sqrt{n^2-6n+8}-(n+6)$$

Answer 1

Divide both the denominator and the numerator by $n^4$, then apply the law of arithmetic operations of limits.

Rigorously speaking, when compute the limit of square roots, if you have not learnt anything about continuous function, then you need to prove them separately. E.G. you need to additionally prove something like $$ \lim_n \sqrt {1 + \frac 1n} = 1 $$ by $\varepsilon$-$\delta$ definition.

UPDATE:

For this limit, you can divide the top and the bottom by $n$ after you do the conjugate part, and the rest are the same as above.

Answer 2

All that matters is the $4n^3$ in the numerator and the $2n^4$ in the denominator. All the other terms are small compared to these.

So the fraction behaves for large $n$ like $\dfrac{4n^3}{2n^4} =\dfrac{2}{n} $ which goes to zero.

Answer 3

For the new question, complete the square under the radical and you have $$\lim_{n \to \infty}\sqrt{(n-3)^2-1}-n-6$$ The intuitive approach is that the $-1$ won't matter as $n$ gets large, so ignore it and you have $-9$

The more formal one is to write $$\lim_{n \to \infty}\sqrt{(n-3)^2-1}-n-6=\lim_{n \to \infty}(n-3)\sqrt{1-\frac 1{(n-3)^2}}-n-6$$ and note the square root goes to $1-\frac 1{2(n-3)^2}$ with the same result.

Answer 4

$\displaystyle \lim_{n\to \infty}\sqrt{n^2-6n+8}-(n+6)=\lim_{n\to \infty}\frac{n^2-6n+8-(n+6)^2}{\sqrt{n^2-6n+8}+(n+6)}=\lim_{n \to \infty}\frac{-18n-28}{\sqrt{n^2-6n+8}+(n+6)}=-9\text{ , dividing by $n$ }$