I must determine the positive constants $c_1$, $c_2$, and $n_0$ such that

$$c_1 n \log(n) \leq 15n \log(5n) \leq c_2 n \log(n)$$

To simplify I attempted to divide by

$$n \log(n))$$

however this yielded a messy result involving changing the base of the logarithm.

  • 1
    $\begingroup$ Recall that $\log (5n) = \log n + \log 5$. $\endgroup$ – Daniel Fischer Sep 5 '16 at 18:07
  • $\begingroup$ Thank you, this helped me solve the problem. $\endgroup$ – Clark Bell Sep 5 '16 at 18:22

$$\begin{align}\lim_{x\to\infty}\frac{15x\log(5x)}{x\log x} &=\lim_{x\to\infty}\frac{15x(\log x+\log 5)}{x\log x}\\ &=\lim_{x\to\infty}\frac{15x\log x+15x\log 5}{x\log x}\\ &=\lim_{x\to\infty}\frac{15x\log x}{x\log x}+\frac{(15\log 5)x}{x\log x} \end{align}$$

Can you take it from here?

  • $\begingroup$ Yes. The answer to the limit above is 15 correct? This is essentially what I did to solve the problem. $\endgroup$ – Clark Bell Sep 5 '16 at 18:31
  • $\begingroup$ Yes, it is. (Filler so I can post this comment) $\endgroup$ – Nathaniel B Sep 6 '16 at 3:04

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