# Help with limits

• Apr 4th 2009, 12:54 AM
omert
Help with limits
I have a problem with my homework:
1. prove only by the defenition of the limit that $\displaystyle \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{1}{{\cos x}} \ne \infty$

• Apr 4th 2009, 03:41 AM
Prove It
Quote:

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

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

If so, it's very easy...

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

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

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

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

$\displaystyle = 0$.

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 $\displaystyle \frac{\sin{x}}{x}\leq \frac{1}{\cos{x}}$.

We just need to find something that $\displaystyle \frac{1}{\cos{x}}$ is less than or equal to...
• Apr 4th 2009, 03:46 AM
mr fantastic
Quote:

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

If so, it's very easy...

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

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

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

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

$\displaystyle = 0$.

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 $\displaystyle \frac{\sin{x}}{x}\leq \frac{1}{\cos{x}}$.

We just need to find something that $\displaystyle \frac{1}{\cos{x}}$ 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 ....
• Apr 4th 2009, 03:59 AM
Prove It
Quote:

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 (Headbang)
• Apr 4th 2009, 04:37 AM
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
• Apr 4th 2009, 04:38 AM
Prove It
Quote:

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...
• Apr 4th 2009, 06:06 AM
matheagle
You need to approach $\displaystyle \pi/2$ from the right and the left.
• Apr 4th 2009, 06:42 AM
Twig
Like matheagle says, approach from left and right.

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

And from the left:

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

I hope this is correct..
• Apr 4th 2009, 06:54 AM
matheagle
you meant....

Quote:

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

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

And from the left:

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

$\displaystyle -\pi/2$ is not $\displaystyle \pi/2^-$

And the limits were backwards.
• Apr 4th 2009, 06:57 AM
Twig
oops yeah, thx (Cool)