Question

lim as x approaches 0 of [1/(3+x)]/x

Answer 1

so, maybe this is not right, but it's the best I've got here. Okay, I went ahead and simplified the expression so \[\lim_{x \rightarrow 0} 1/(x^2+3x)\]
Then I went ahead and completed the square in the denominator. So we have\[L=1/(x^2 +3x + 9/4) -[1/(9/4)]\]
this simplifies to\[L=1/(x+3/2)^2 -(4/9)\]
Then, subbing in 0, the limit =0.

Answer 2

Sweetness, thanks :D

Answer 3

I guess to make the algebra easier you could get (x^2+3x)= 1/L, them complete squares, then get L back, but it all solves the same