Determine if the following sequence converge and if so, find the limit:

$$\left(\frac{\ n^{2}+17n-5}{16n^{3}\sec (n)-7}\right)_{n=0}^\infty$$

Basically i wrote the limit of the sequence as x approaches $\infty$ and then I substituted $\sec(x)$ with $\frac{1}{cos (x)}$ then: $$\lim_{x\to \infty}\frac{\cos (x) ( x^{2}+17x-5)}{16x^{3}-7cos (x)}$$

For $x^{+}$ , $16x^{3}$ will be dominant, therefore $-7\cos(x)$ won't make a significant difference.

$16x^{3} - 7cos(x)$ $\approx$ $16x^{3}$

\begin{align} \lim_{x\to \infty}\frac{(x^2 +17x-5) \cos x }{16x^3} & = \lim_{x\to \infty} \left(\frac{x^2 +17x-5}{16x^3}\right)\cos x \\ &= \lim_{x\to \infty} \left( \frac{\ 1}{16x}+ \frac{17}{16x^{2}}- \frac {5}{16x^{3}}\right)\cos x\\ \end{align}

All of the terms above equal $0$ since anything $\infty = 0$ therefore the limit is $0$.

Since the limit exists,

$$\left(\frac{\ n^{2}+17n-5}{16\sec (n)-7}\right)_{n=0}^\infty\;\;\;\;\; \text{converges to $0$}$$

Is this correct and complete or should I use the squeeze theorem?

And if so, how to apply it? Thanks in advance!

  • $\begingroup$ Where did $16x^3$ come from? $\endgroup$
    – user258700
    Mar 19 '16 at 21:00
  • $\begingroup$ Is your denominator $16\sec(n)-7$ or $16n^3\sec(n)-7$? You start with the former, but seem to wind up with the latter. $\endgroup$ Mar 19 '16 at 21:02
  • $\begingroup$ i'm sorry, i spelled wrong, it s $16n^{3}sec (n)$ - 7 I corrected it now. $\endgroup$
    – Debbie
    Mar 19 '16 at 21:11

The expression equals

$$\tag 1\frac{1+17/n -5/n^2}{16n\sec n -7/n^2}.$$

The numerator $\to 1,$ clearly. What about the denominator? In absolute value the denominator is

$$\ge |16n\sec n| - |7/n^2| \ge 16n -7.$$

So for $n$ large, the absolute value of $(1)$ is less than or equal to


The limit in question is thus $0.$


You can use squeeze theorem by noting $\mid \text{sec } n\mid \geq 1$. So it allows you to squeeze the function by using denominator.


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