# proving $\lim\limits_{x\to \frac{\pi }{2}}\frac1{\cos x}\ne \infty$

I have to prove the following equation for homework $$\lim_{x\to \frac{\pi }{2}}\frac1{\cos x}\ne \infty$$

The proof must be done by proving that exists a $M>0$ for which for every $l>0$ exists an $x$ so that $0<|x-(\pi/2)|<l$ but $\lim\limits_{x\to \frac{\pi }{2}}\frac1{\cos x}\le m$

I can't seem to figure this one out.

I would greatly appropriate anyone who tries to help me out :) Thanks

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Did you intend capital $M$ and lower-case $m$ to be understood as being the same? You shouldn't do that, since it's standard to sometimes use capital and lower-case letters to refer to different things in the same problem. – Michael Hardy Nov 18 '11 at 19:20
My (late) hint would be: first try to understand that $\lim_{x\to 0}\frac{1}{x}\neq\infty$. – wildildildlife Nov 25 '11 at 23:02

First we use the geometry to find out what is really going on. Then we will be ready to go to the $M$ and $l$ stuff. Very informally, if $\lim_{x\to \pi/2} \frac{1}{\cos x} =\infty$, that would mean that if $x$ is close but not equal to $\frac{\pi}{2}$ then $\frac{1}{\cos x}$ is big positive.

If $x$ is a tiny bit below $\frac{\pi}{2}$, then $\frac{1}{\cos x}$ is indeed big positive, since $\cos x$ is positive and close to $0$. But if $x$ is a bit bigger than $\frac{\pi}{2}$, then $\cos x$ is close to $0$ but negative, so $\frac{1}{\cos x}$ is certainly not big positive. So the bad guys are the $x$ that are a bit bigger than $\frac{\pi}{2}$.

Now that we know what's going on, let's write a formal proof. I will try to use the symbols that you used.

What would it mean for the limit to be $\infty$? To save typing, and for the sake of generality, I will write for a while $a$ for $\frac{\pi}{2}$ and $f(x)$ for $\frac{1}{\cos x}$.

We have $\lim_{x\to a}f(x)=\infty$ if for any $M$, there is an $l>0$ such that for any $x$ such that $|x-a|<l$, then $f(x)>M$. You can think of it this way. Suppose we are given a specific big $M$, like $1000$, or $99999999$. We want to be able to exhibit a number $l$ such that if the distance $|x-a|$ of $x$ from $a$ is $<l$, then $f(x)$ is guaranteed to be $>M$.

Think of the number $M$ as a challenge, and of $l$ as a response. To the challenge "ensure that $f(x)>M$" we must always be able to come up with an appropriate response "if $|x-a|<l$, then I can guarantee that $f(x)>M$." (The appropriate response $l$ will in general depend on $M$.)

The rest is easy. Let $M=17$. Can we come up with an $l$ such that if $\left|x-\frac{\pi}{2}\right|<l$, then $\frac{1}{\cos x}>17$? Definitely not. For whatever $l$ we pick, there will be a number $x$ such that $\frac{\pi}{2}<x<\pi$ and $\left|x-\frac{\pi}{2}\right|<l$. And at any such number $x$, the number $\frac{1}{\cos x}$ will be negative, so definitely not bigger than $17$. So for the challenge $M=17$, there is no possible response $l$.

We conclude that it is not true that $\displaystyle\lim_{x\to \pi/2} \frac{1}{\cos x} =\infty$. Indeed, a small variant of the argument shows that the limit in this case does not exist.

Comment: You wrote $\displaystyle\lim_{x\to \pi/2} \frac{1}{\cos x} \ne \infty$. I prefer not to do that, for writing it that way may carry the impression that the limit exists, but happens to be something other than $\infty$.

With some practice, this $M$, $l$ business, and related ideas, will become clearer to you. It is a genuinely subtle idea, and takes time to become fully absorbed.

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Thanks - Awesome answer :) – Jason Nov 25 '11 at 18:42

If $x$ is slightly bigger than $\pi/2$, then $\cos x$ is negative.

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Actually, the limit is $\infty$ if instead of gluing a $+\infty$ onto one end of the real line and $-\infty$ onto the other, you just add a single $\infty$ at both ends, making the line topologically a circle. In trigonometry and when dealing with rational functions, that makes sense. – Michael Hardy Nov 18 '11 at 18:05
Would that be a "wheel"? – Pedro Tamaroff Apr 25 '12 at 23:18

You need to prove that $$\lim_{x\to \frac{\pi }{2}^+}\frac1{\cos x} \neq \lim_{x\to \frac{\pi }{2}^-}\frac1{\cos x}.$$

This is trivial since for $x>\frac{\pi}{2}$, we have

$$\frac{1}{\cos x}\leq -1,$$

and similarly for $x<\frac{\pi}{2}$, we have

$$\frac{1}{\cos x}\geq 1.$$

This means $$\lim_{x\to \frac{\pi }{2}}\frac1{\cos x}$$ just do not exist.

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I think you mean $$\lim_{x\to\frac{\pi}{2}^+}\frac1{\cos x}\neq\lim_{x\to\frac{\pi}{2}^-}\frac1{\cos x}$$ – robjohn Nov 18 '11 at 18:25
Oh, yeah, right :) – Santa Zhang Nov 18 '11 at 18:27
@SantaZhang, You are correct this is a great answer but I am limited to proving this by the method stated in the question :) – Jason Nov 18 '11 at 18:31
@Jason: If you are limited to showing that there "exists a $M>0$ for which for every $l>0$ exists an $x$ so that $0<|x-(\pi/2)|<l$ but $\lim\limits_{x\to \frac{\pi }{2}}\frac1{\cos x}\le m$", then you will not find a solution. This statement assumes that a finite limit exists. Unfortunately, as all the answers show, there is no limit because no matter how small we choose $\delta>0$, there are $x$ with $|x-\frac{\pi}{2}|<\delta$ so that $\frac{1}{\cos(x)}<-1$ and $x$ with $|x-\frac{\pi}{2}|<\delta$ so that $\frac{1}{\cos(x)}>+1$. – robjohn Nov 25 '11 at 19:46