1. ## Help with limits

I have a problem with my homework:
1. prove only by the defenition of the limit that $
\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{1}{{\cos x}} \ne \infty

$

2. Originally Posted by omert
I have a problem with my homework:
1. prove only by the defenition of the limit that $
\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{1}{{\cos x}} \ne \infty

$

Are you allowed to use L'Hospital's Rule?

If so, it's very easy...

$\lim_{x \to \frac{\pi}{2}} \frac{1}{\cos{x}} = \lim_{x \to \frac{\pi}{2}} \frac{\frac{d}{dx}(1)}{\frac{d}{dx}(\cos{x})}$

$= \lim_{x \to \frac{\pi}{2}} \frac{0}{-\sin{x}}$

$= \frac{0}{-\sin{\frac{\pi}{2}}}$

$= \frac{0}{-1}$

$= 0$.

Without L'Hospital, not so easy...

All I can think of is the Sandwich Theorem, but I can't think of what to sandwich it inside.

I know that $\frac{\sin{x}}{x}\leq \frac{1}{\cos{x}}$.

We just need to find something that $\frac{1}{\cos{x}}$ is less than or equal to...

3. Originally Posted by Prove It
Are you allowed to use L'Hospital's Rule?

If so, it's very easy...

$\lim_{x \to \frac{\pi}{2}} \frac{1}{\cos{x}} = \lim_{x \to \frac{\pi}{2}} \frac{\frac{d}{dx}(1)}{\frac{d}{dx}(\cos{x})}$

$= \lim_{x \to \frac{\pi}{2}} \frac{0}{-\sin{x}}$

$= \frac{0}{-\sin{\frac{\pi}{2}}}$

$= \frac{0}{-1}$

$= 0$.

Without L'Hospital, not so easy...

All I can think of is the Sandwich Theorem, but I can't think of what to sandwich it inside.

I know that $\frac{\sin{x}}{x}\leq \frac{1}{\cos{x}}$.

We just need to find something that $\frac{1}{\cos{x}}$ is less than or equal to...
The given limit is not of indeterminant form, l'Hopitals rule is not valid. The limit is not equal to zero.

The limit does not exist. My suggestion is to consider the left hand and right hand limits ....

4. Originally Posted by mr fantastic
The given limit is not of indeterminant form, l'Hopitals rule is not valid. The limit is not equal to zero.

The limit does not exist. My suggestion is to consider the left hand and right hand limits ....
Oops, my mistake... I knew I was missing something obvious

5. I can't use any theorem. I need to prove it only by the defenition. let be M>0 so the any delta>0.... x-pi/2 -->>> 1/cosx<=M

6. Originally Posted by omert
I can't use any theorem. I need to prove it only by the defenition. let be M>0 so the any delta>0.... x-pi/2 -->>> 1/cosx<=M
You could just look at the graph...

7. You need to approach $\pi/2$ from the right and the left.

8. Like matheagle says, approach from left and right.

$\lim_{x\rightarrow \frac{\pi}{2}} \frac{1}{cos x} = \infty$

And from the left:

$\lim_{x\rightarrow \frac{-\pi}{2}} \frac{1}{cos x} = -\infty$

I hope this is correct..

9. you meant....

Originally Posted by Twig
Like matheagle says, approach from left and right.

$\lim_{x\rightarrow \pi/2^+} \frac{1}{cos x} = -\infty$

And from the left:

$\lim_{x\rightarrow \pi/2^-} \frac{1}{cos x} = \infty$
$-\pi/2$ is not $\pi/2^-$

And the limits were backwards.

10. oops yeah, thx