Which of the following limits equal infinity ( but not + infinity or - infinity)
a) lim (x->0) sinx/x^2
b) lim (x->3) (x+3) / (x^2 - 9)
Im not sure how to determine if it's +inf., -inf., or just infinity
Please help....Thanks.
Hello jzellt!
Did you ever hear about the Rule of L'Hospital?
a) lim (x->0) sinx/x^2
lim (x->0) sinx = 0 and lim (x->0) x^2 = 0
So [sin(x)]' = cos(x) and [x^2]' = 2x
Thus lim (x->0) sinx/x^2 = lim (x->0) cosx/(2x)
It is lim (x->0) cosx = 1 and lim(x->0)2x = 0
So
lim (x->0) cosx/(2x) = $\displaystyle \pm \infty$
b) lim (x->3) (x+3) / (x^2 - 9)
You want to find the limes of +3? You are sure about that? In that case it is
lim (x->3) (x+3) / (x^2 - 9) = $\displaystyle \pm \infty$
because (x+3) -> 6, if x->3
and x^2-9 -> 0, if x -> 0
Regards
Rapha