What is : lim t =0 [ (sqrt (t^2+9) - 3) / t^2]

The value that we have to determine
is:

lim
t-->0[sqrt(t^2+9)-3/t^2]

If we substitute t = 0, we get the
indeterminate form 0/0, this allows us to use l'Hopital's rule and substitute the numerator and
denominator with their derivatives.

=> lim t-->0 [
2*t*(1/2)/sqrt(t^2 + 9)*2*t]

=> lim t-->0
[1/2*sqrt(t^2 + 9)]

substitute t =
0

=> (1/2*sqrt 9)

=>
1/6

The required value of the limit is
1/6