is my proving limit by epsilon-delta defnition ok?

[orir]

i need to prove that \(\displaystyle \displaystyle{\lim _{x \to {pi/2}}}1/cos(x) != infinity\)
so in other words - i need to prove that there's a positive number \(\displaystyle M \) (as big as we want) so for every \(\displaystyle delta>0\) there's an \(\displaystyle X\) for it \(\displaystyle 0<|x-(pi/2)|<delta\)
so \(\displaystyle 1/cos(x)<=M\)

the proof:
let \(\displaystyle lamda>0\). let\(\displaystyle M=20\), and \(\displaystyle x=1.5\)
we can see that \(\displaystyle 1/cos(1/5)<20\)

is it enough?

 

[Nehushtan]

i need to prove that \(\displaystyle \displaystyle{\lim _{x \to {\frac{\pi}{2}}}}\frac{1}{\cos x}\ne\infty\)
so in other words - i need to prove that there's a positive number \(\displaystyle M \) (as big as we want) so for every \(\displaystyle \delta>0\) there's an \(\displaystyle x\) for it \(\displaystyle 0<\left|x-\frac{\pi}{2}\right|<\delta\) so \(\displaystyle \frac{1}{\cos x}\leq M\)

I have improved your LaTeX presentation. Please learn to format your equations properly; I misread your question initially and thought you wanted to prove something else. :?

The limit does exist if you let \(\displaystyle x\) approach \(\displaystyle \frac{\pi}2\) from the left only. So you need to take \(\displaystyle x>\frac{\pi}2\) to show that \(\displaystyle \frac1{\cos x}\) does not tend to infinity as \(\displaystyle x\) approaches from the right. Hint: Let \(\displaystyle M=1\) (any positive number will do) and given any \(\displaystyle \delta>0\), take \(\displaystyle x\) in the interval \(\displaystyle \left(\frac{\pi}2,\,\frac{\pi}2+\delta'\right)\) where \(\displaystyle \delta'=\min\left\{\delta,\pi\right\}\). In other words, always take \(\displaystyle x\) so that \(\displaystyle \frac1{\cos x}\) is negative and so less than \(\displaystyle M\).