- #1
Physicist_FTW
- 9
- 0
use series expansion to fine the limit as x--->0 of
1/(sin^2)(x) - 1/x^2
Sin^2(x)=(X-X^3/3!)^2 I've assumed this gave me
Sin^2(x)=X^2-2X^4/6+X^6/36
flip this over is that equivalent to 1/Sin^2(x)?
1/(sin^2)(x) - 1/x^2
Sin^2(x)=(X-X^3/3!)^2 I've assumed this gave me
Sin^2(x)=X^2-2X^4/6+X^6/36
flip this over is that equivalent to 1/Sin^2(x)?