# Thread: Can someone explain why this is?

1. ## 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?

2. 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.

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

Graph

... which can produce the graph of $\displaystyle y = \frac{1}{\sin^2{x}} = csc^2{x}$

4. Thanks, it makes a lot more sense now.

5. 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.

6. Yes, I already understand that a negative number squared becomes positive.