# Limit of $\lim \limits_{x \to \frac{5π}{2}^+} \frac{5x - \tan x}{\cos x}$

So I have the following problem:

$$\lim \limits_{x \to \frac{5π}{2}^+} \frac{5x - \tan x}{\cos x}$$

I can't figure out how to get the limit. I tried splitting it up to:

$$\lim \limits_{x \to \frac{5π}{2}^+} \Big(\frac{5x}{\cos x} - \frac{\tan x}{\cos x}\Big)$$

I'm lost and unsure of what to do next. I'm taking a Calc 1 class and we have not yet gotten to L'hopitals and other methods yet (and also I am not sure how I could incorporate those ideas either).

We see that the numerator $5x-\tan(x)$ diverges to $\infty$, and that the denominator $\cos(x)$ goes to $0$. So, we're taking something that grows large and dividing it by something that becomes very small, and thus the limit diverges to $\infty$.

• Thank you! Your explanation makes sense to me, but how would I show that algebraically? – Antor Paul Sep 14 '16 at 3:14
• @AntorPaul Well, since $\tan(x) = \frac{\sin(x)}{\cos(x)}$, all you really need is that $\sin\left(\frac{5\pi}{2}\right) = 1$ and $\cos\left(\frac{5\pi}{2}\right) = 0$. – Carl Schildkraut Sep 14 '16 at 3:18
• the limit is actually -∞ but that requires some analysis to determine. – Jack Tiger Lam Sep 14 '16 at 4:00

Hint:- $$\lim_{x\to \frac{5\pi}{2}}\frac{5x-\tan x}{\cos x}=\lim_{x\to \frac{5\pi}{2}}\frac{5x-\tan x}{\sin (\frac{\pi}{2}-x)}=\lim_{x\to \frac{5\pi}{2}}\frac{5x-\tan x}{\sin(\frac{5\pi}{2}-x)}$$ Now divide denominator and numerator by $\frac{5\pi}{2}-x$

• How do you do it ? – Claude Leibovici Sep 14 '16 at 7:36
• Since $\sin$ wave is periodic with period $2\pi$, so $\sin(\frac{\pi}{2}-x)=\sin(2\pi+\frac{\pi}{2}-x)$ – Mayank Deora Sep 15 '16 at 2:12

This is not an answer but it is too long for a comment.

The question which could have been asked is : how does approach the expression when $x \to \frac{5\pi}{2}$ ?

One solution could be to first change variable $x=y+\frac{5\pi}{2}$ which makes $$A=(5 x-\tan (x)) \sec (x)=-\frac{1}{2} (10 y+2 \cot (y)+25 \pi ) \csc (y)$$ and now use Taylor series around $y=0$ $$\cot(y)=\frac{1}{y}-\frac{y}{3}+O\left(y^2\right)$$ $$\csc(y)=\frac{1}{y}+\frac{y}{6}+O\left(y^2\right)$$ which make $$A=-\frac{1}{y^2}-\frac{25 \pi }{2 y}-\frac{29}{6}+O\left(y^1\right)$$ which shows the limit $(-\infty)$ and how it is approached.

To show how "good" is the approximation, let me use $y=\frac{ \pi }{6}$. The exact value is $$A=-2 \sqrt{3}-\frac{80 \pi }{3}\approx -87.24$$ while the above approximation would give $$-\frac{479}{6}-\frac{36}{\pi ^2}\approx -83.48$$.