Help with limits

**omert** · April 4th 2009, 12:54 AM

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 $

Thanks ahead

**Prove It** · April 4th 2009, 03:41 AM

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 $

Thanks ahead

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...

**mr fantastic** · April 4th 2009, 03:46 AM

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 ....

**Prove It** · April 4th 2009, 03:59 AM

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

**omert** · April 4th 2009, 04:37 AM

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
thanks ahead

**Prove It** · April 4th 2009, 04:38 AM

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
thanks ahead

You could just look at the graph...

**matheagle** · April 4th 2009, 06:06 AM

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

**Twig** · April 4th 2009, 06:42 AM

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..

**matheagle** · April 4th 2009, 06:54 AM

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.

**Twig** · April 4th 2009, 06:57 AM

oops yeah, thx

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