prove that $(n+3)^{2}=\Theta(n-5)^{2}$ I need to find $c_1,c_2,n_0&gt;0$ so for all $n\geq n_0$:
$$c_{1}(n-5)^{2}\leq(n+3)^{2}\leq c_{2}(n-5)^{2}$$
I have problem with the right side of the inequality. I can't find c2.
thanks for any help :)
Write
$$\sqrt{c_1}\le\frac{n+3}{n-5}\le\sqrt{c_2}$$
or
$$\sqrt{c_1}\le1+\frac8{n-5}\le\sqrt{c_2}.$$
Now choose some $n_0&gt;5$.
Write
$$\sqrt{c_1}\le\frac{n+3}{n-5}\le\sqrt{c_2}$$
or
$$\sqrt{c_1}\le1+\frac8{n-5}\le\sqrt{c_2}.$$
Now choose some $n_0&gt;5$.