To solve we will simplify the numerator:

sqrt(t^2+9) - 3 / t^2

We will multiply and divide by (sqrt(t^2+9) + 3)

(sqrt(t^2+9) -3)(sqrt(t^2+9) +3) / t^2(sqrt(t^2+9) +3)

==> (t^2 +9 -9) / t^2 ( sqrt(t^2+9) +3)

==> t^2/t^2 ( sqrt9t^2+9) +3)

==> 1/(sqrt(t^2+9) + 3)

Now we will substitute with t= 0 to find the limit.

==> lim ==> 0

==> 1/(sqrt9 + 3)

==> 1/ (3+3) = 1/6

**Then the limit is 1/6**

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**

thanks. My teacher is not allowed to use l'Hopital's. I need to use factoring or rationalization method. thanks