Question

evaluate
lim(x->0) (sin^2x -x^2)/(x^2* sin^2x)

Answer 1

yea way too early for this one

Answer 2

this tends to involve trying to finagle a sin(x)/x into the picture

Answer 3

if we sub x=0 the function = 0/0 indeterminate
so i think we need to use hopital's rule

Answer 4

i must admit i'm pretty rusty at limits

Answer 5

\[\lim_{(x->0)} \frac{sin^2x -x^2}{x^2\ sin^2x}\]
\[\lim_{(x->0)} \frac{(sin(x) -x)(sin(x) +x)}{x\ sin(x)x\ sin(x)}\]
\[\lim_{(x->0)} \frac{sin(x) +x}{x\ sin(x)}*\frac{sin(x) -x}{x\ sin(x)}\]
\[\lim_{(x->0)} \frac{sin(x)}{x\ sin(x)}+\frac{x}{x\ sin(x)}*\frac{sin(x)}{x\ sin(x)}-\frac{x}{x\ sin(x)}\]
\[\lim_{(x->0)} \frac{1}{x}+\frac{1}{sin(x)}*\frac{1}{x}-\frac{1}{sin(x)}\]

apparently i cant recall a nice way to do it this morning :)

Answer 6

its not coming by l'hospital...

Answer 7

ugh..wolfram maybe?

Answer 8

http://www4d.wolframalpha.com/Calculate/MSP/MSP140119h4bf22d81d1h2100003ihaei7e167cf742?MSPStoreType=image/gif&s=59&w=488&h=2528

Pretty Long Solution .....Need a better Way........

*BookMark*

Answer 9

http://www.wolframalpha.com/input/?i=limit++%28-x%5E2%2BSin%5Bx%5D%5E2%29%2F%28x%5E2+Sin%5Bx%5D%5E2%29+as+x+-%3E0