0
$\begingroup$

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).

$\endgroup$
0
$\begingroup$

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 $\infty$.

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

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 $\frac{5\pi}{2}-x$

$\endgroup$
  • $\begingroup$ How do you do it ? $\endgroup$ – Claude Leibovici Sep 14 '16 at 7:36
  • $\begingroup$ Since $\sin$ wave is periodic with period $2\pi$, so $\sin(\frac{\pi}{2}-x)=\sin(2\pi+\frac{\pi}{2}-x)$ $\endgroup$ – Mayank Deora Sep 15 '16 at 2:12
0
$\begingroup$

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 $$-\frac{479}{6}-\frac{36}{\pi ^2}\approx -83.48$$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.