Multiply the numerator and denominator by [tex]\sqrt{t^2+9}+3[/tex]. The motivation for this is to exploit the difference of squares identity,
[tex]a^2-b^2=(a-b)(a+b)[/tex]
In this case, [tex]a=\sqrt{t^2+9}[/tex] and [tex]b=3[/tex]. So we have
[tex]\dfrac{\sqrt{t^2+9}-3}{t^2}=\dfrac{\sqrt{t^2+9}-3}{t^2}\cdot\dfrac{\sqrt{t^2+9}+3}{\sqrt{t^2+9}+3}[/tex]
[tex]\dfrac{\sqrt{t^2+9}-3}{t^2}=\dfrac{(\sqrt{t^2+9})^2-3^2}{t^2(\sqrt{t^2+9}+3)}[/tex]
[tex]\dfrac{\sqrt{t^2+9}-3}{t^2}=\dfrac{(t^2+9)-9}{t^2(\sqrt{t^2+9}+3)}[/tex]
[tex]\dfrac{\sqrt{t^2+9}-3}{t^2}=\dfrac{t^2}{t^2(\sqrt{t^2+9}+3)}[/tex]
[tex]\dfrac{\sqrt{t^2+9}-3}{t^2}=\dfrac1{\sqrt{t^2+9}+3}[/tex]
Now the expression in the limit is continuous at [tex]t=0[/tex], so the limit can be computed by substitution:
[tex]\displaystyle\lim_{t\to0}\frac{\sqrt{t^2+9}-3}{t^2}=\lim_{t\to0}\frac1{\sqrt{t^2+9}+3}=\frac1{\sqrt{9}+3}=\boxed{\frac16}[/tex]
