<s>[INST] Find the following limit 0/0 $$\lim_{t\to 0} \frac{\left(\sqrt{t+9}-3 \sqrt[3]{2 t+1}\right)}{\sqrt[3]{t+8}-2 \sqrt[3]{3 t+1}}$$
I tried to multiply the nominator by conjugation but got even bigger expression.
 [/INST] Let $a=\sqrt{t+9}$, $b=3\sqrt[3]{2t+1}$, $c=\sqrt[3]{t+8}$ and $d=2\sqrt[3]{3t+1}$.
We have 
$$ \lim_{t\to 0} \frac{a-b}{c-d} = \lim_{t\to 0}\left(\frac{a^6 -b^6}{c^3-d^3}\right)\left( \frac{c^2+cd + d^2}{a^5+a^4b+a^3b^2+a^2b^3+ab^4+b^5}\right)
= \lim_{t\to 0}\left(\frac{(t+9)^3 -(3^6(2t+1)^2)}{t+8 - 8(3t+1)}\right)\left( \frac{c^2+cd + d^2}{a^5+a^4b+a^3b^2+a^2b^3+ab^4+b^5}\right)=\left(\frac{3(9^2) -4(3^6)}{-23}\right)\left(\frac{3(4)}{6(3^5)}\right)=\frac{22}{23}.
$$
Note that on the last line we used: $$\lim_{t\to 0}\frac{(t+9)^3 -(3^6(2t+1)^2)}{t+8 - 8(3t+1)} = \lim_{t\to 0}\frac{(t^3+3(9)t^2+3(9)^2t + 9^3) - (3^6(4t^2+4t+1))}{-23t} = \lim_{t\to 0}\frac{(t^3+3(9)t^2+3(9)^2t) - (3^6(4t^2+4t))}{-23t} = \lim_{t\to 0}\frac{(t^2+3(9)t^1+3(9)^2) - (3^6(4t+4))}{-23} = \frac{3(9^2) -4(3^6)}{-23}$$
and $\lim_{t\to 0} a = \lim_{t\to 0} b=3$ and $\lim_{t\to 0} c = \lim_{t\to 0} d=2$.
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