Find parameters in limit

Question

Find real a,b so that $$ \lim_{n\to\infty} \sqrt{n^2-n+1}-an-b =-\infty $$ I have no idea what to do. Thank you.

Answer 1

You can rewrite this as $$ n\left (\sqrt{1 - \frac{1}{n} + \frac{1}{n^2}} - a\right) - b $$ As $n\to \infty$, the only term that really matters is the first one.

Hence, $b$ can be any real number.

Now the term under the square root gets very close to 1, and if the limit has to be $-\infty$, then the term $$ t_n := \sqrt{1 - \frac{1}{n} + \frac{1}{n^2}} - a $$ must be negative at some point. Note that $$ \lim_{n\to \infty} t_n = 1 - a $$ Hence, if $1-a < 0$, then $t_n < 0$ for large $n$.

Hence, for the limit to be $-\infty$, it must happen that $a>1$

Answer 2

Putting $n=\frac1h,$

$$A=\lim_{n\to\infty}\sqrt{n^2-n+1}-an-b=\lim_{h\to0}\frac{(1+h^2-h)^{\frac12}-a-bh}h$$

(Using Generalized Binomial Expansion)

$$A=\lim_{h\to0}\frac{1+\frac{h^2-h}2+O(h^2)-a-bh}h$$

$$=\lim_{h\to0}\left(\frac{1-a}h+O(h)\right)-b-\frac12$$

So, we need $1-a<0$

Answer 3

Consider how the square root with everything inside it tends to approach $n$ as $n$ goes to infinity. Thus, $$\lim_{n\to\infty} n - an - b \Longrightarrow n(1 - a) - b = -\infty$$

for what values of $n$?

Answer 4

The parameter $b$ is irrelevant: if the limit is inexistent/finite/$\infty$/$-\infty$ without it it will be in the same class with it.

What happens when $a\le0$?

And when $a>0$, try considering $$ \sqrt{n^2-n+1}-an= \frac{(\sqrt{n^2-n+1}-an)(\sqrt{n^2-n+1}+an)}{\sqrt{n^2-n+1}+an}= \frac{n^2-n+1-a^2n^2}{\sqrt{n^2-n+1}+an} $$

Answer 5

Factor an $n$ out of the expression as follows: $$n\left(\sqrt{1-\frac{1}{n}+\frac{1}{n^2}}-a-\frac{b}{n}\right)$$ Now you can see that the remaining square root approaches 1, and the $\frac{b}{n}$ term approaches $0$. If $a>1$, then the expression in the parentheses is negative for large $n$, so the limit is $-\infty$. If $a<1$ then the expression is positive for large $n$, so the limit is $+\infty$. In both cases $b$ can be anything.