Question

1. (10 points) evaluate \\( \lim_{t \to 0} \frac{\sqrt{t^2 + 9} - 3}{t^2} \\).

Explanation:

Step1: Rationalize numerator

Multiply by $\frac{\sqrt{t^2+9}+3}{\sqrt{t^2+9}+3}$:
$\lim_{t
ightarrow 0}\frac{(\sqrt{t^2+9}-3)(\sqrt{t^2+9}+3)}{t^2(\sqrt{t^2+9}+3)}$

Step2: Simplify numerator

Use $(a-b)(a+b)=a^2-b^2$:
$\lim_{t
ightarrow 0}\frac{(t^2+9)-9}{t^2(\sqrt{t^2+9}+3)}=\lim_{t
ightarrow 0}\frac{t^2}{t^2(\sqrt{t^2+9}+3)}$

Step3: Cancel $t^2$

$\lim_{t
ightarrow 0}\frac{1}{\sqrt{t^2+9}+3}$

Step4: Substitute $t=0$

$\frac{1}{\sqrt{0+9}+3}=\frac{1}{3+3}=\frac{1}{6}$

Answer:

$\frac{1}{6}$