In case you havent solved these yet (and because i love easy math!)

i solved them for you:

1)

limx->0 (cscx-cotx)

= 1/sinx - 1/tanx

= 1/sinx - cosx/sinx

= (1-cosx)/sinx

= (1-cos^2x)/sinx * 1/(1+cosx) (multiplied by (1+cosx)/(1+cosx))

= sin^2x/sinx*1/1+cosx

= sinx * 1/1+cosx

= 0

2)

limx->inf (1+3/x+5/x^2)^x

= limx->inf e^(x log(1+3/x+5/x^2)) <- call this equation 1

im just going to look at the exponent part.

xlog( 1+3/x+5/x^2)

= x( log( 1/x^2(x^2+3x+5))

= x( log(1/x^2) + log(x^2+3x+5) )

= x( log(x^2+3x+5) - log(x^2) )

as x->inf

log(x^2+3x+5) - log(x^2) = 0

thus going back to equation 1

e^0 = 1 which is the answer..

any questions? i dont like using l'hopitals rule.. its the pansy way to solve limits!

hopefully im right on this!