# Can someone explain why this is?

• Sep 5th 2010, 07:28 AM
ImaCowOK
Can someone explain why this is?
Can someone explain why the limit of the function 1/sin²x is ∞ as x approaches 0(from both left and right).

I'm just starting to learn this, so as simple an explanation as you can manage would be appreciated.

What does the graph of 1/sin² x look like?
• Sep 5th 2010, 07:39 AM
Jester
Since $\displaystyle \lim_{x \to 0} \sin^2 x = 0.$ This means we can make $\sin^2 x$ as close to zero as we wish by making $x$ sufficiently close to $0$. Thus we can make $\dfrac{1}{\sin^2x}$ as large as we wish by making $x$ sufficiently close to $0$.

We can be a little more precise with a $\delta-\epsilon$ definition if you wish.
• Sep 5th 2010, 08:04 AM
skeeter
Quote:

Originally Posted by ImaCowOK
...What does the graph of 1/sin² x look like?

Graph

... which can produce the graph of $\displaystyle y = \frac{1}{\sin^2{x}} = csc^2{x}$
• Sep 5th 2010, 10:00 AM
ImaCowOK
Thanks, it makes a lot more sense now.
• Sep 5th 2010, 06:07 PM
HallsofIvy
Note that for x= 0.001, sin(x)= 0.00001745329, approximately, so that [tex]sin^2(x)= 0.000000000304617419755778[tex] and then $\frac{1}{sin^2(x)}= 3282806350$.

Of course, if x= -0.001, sin(x)= -0.0000174329 and so $sin^2(x)$ and $\frac{1}{sin^2(x)}$ are exactly the same as above.
• Sep 5th 2010, 07:38 PM
ImaCowOK
Yes, I already understand that a negative number squared becomes positive.