How to show that pairs of functions are of the same order? If we have these pairs of functions, how can we show that they are of the same
order?
a) $3x + 7,\quad x$
b) $2x^2 + x − 7,\quad x^2$
c) $\lfloor x + 1/2\rfloor ,\quad x$
d) $\log(x^2 + 1),\quad \log_2 x$
e) $\log_{10} x,\quad \log_2 x$
Thanks guys!
 A: You want to show that $\dfrac f g \to K$ as $x\to\infty$, where $K\neq 0$ is some constant. For polynomials, this is almost immediate. For the logarithms, you can use L'Hôpital.
A: My interpretation of the claim that $f$ and $g$ are of the same order is that $f=\Theta(g)$ as $x\to\infty$. By definition this means that there are $x_0,c_1,c_2>0$ such that
$$c_1g(x)\le f(x)\le c_2g(x)\quad\text{for all }x\ge x_0\;.$$
Of course this is equivalent to
$$c_1\le\frac{f(x)}{g(x)}\le c_2\quad\text{for all }x\ge x_0\;.$$
This is weaker than saying that $\dfrac{f(x)}{g(x)}$ approaches a positive limit as $x\to\infty$, as may be seen by considering the functions $f(x)=100+\sin x$ and $g(x)=100$. However, the pairs in the question all meet the stronger requirement, though it may take a moment to see this in for (c):
$$x\le\left\lfloor x+\frac12\right\rfloor\le x+1\;,$$
so
$$1\le\frac{\left\lfloor x+\frac12\right\rfloor}x\le1+\frac1x\;,$$
and $$\lim_{x\to\infty}\frac{\left\lfloor x+\frac12\right\rfloor}x=1\;.$$
For the pairs involving logarithms, remember that $\log_ax=\dfrac{\log_bx}{\log_ba}$.
