Find the following limit 0/0

Question

$$\lim_{t\to 0} \frac{\left(\sqrt{t+9}-3 \sqrt[3]{2 t+1}\right)}{\sqrt[3]{t+8}-2 \sqrt[3]{3 t+1}}$$

I tried to multiply the nominator by conjugation but got even bigger expression.

Answer 1

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 $\lim_{t\to 0} c = \lim_{t\to 0} d=2$.

Answer 2

Use $\lim_{x\to 0}\frac{(1+x)^a-1}{x}=a$ and $\frac1{a^{1/3}-b^{1/3}}=\frac{a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}}{a-b}$

Then the limit equals to \begin{align} & \lim_{t\to 0}3[(1+t/9)^{1/2}-(1+2t)^{1/3}] \times \frac{(t+8)^{2/3}+(t+8)^{1/3}(24t+8)^{1/3}+(24t+8)^{2/3}}{(t+8)-(24t+8)} \\ = & -\frac{36}{23}\lim_{t\to 0}\frac{[(1+t/9)^{1/2}-1]-[(1+2t)^{1/3}-1]}{t} \\ = & -\frac{36}{23}\left(\lim_{t\to 0}\frac1{9}\frac{[(1+t/9)^{1/2}-1]}{t/9}-\lim_{t\to 0}\frac{2[(1+2t)^{1/3}-1]}{2t}\right) \\ = & -\frac{36}{23}(\frac{1}{18}-\frac{2}{3}) \\ = & \frac{22}{23}. \end{align}

Answer 3

The simplest (and systematic) way is to use Taylor's formula at order $1$: \begin{align*} \frac{\left(\sqrt{t+9}-3 \sqrt[3]{2 t+1}\right)}{\sqrt[3]{t+8}-2 \sqrt[3]{3 t+1}}&=\frac{3\biggl(\sqrt{1+\dfrac t9}-\sqrt[3]{1+2t}\biggr)}{2\biggl(\sqrt[3]{1+\dfrac t8}-\sqrt[3]{1+3t}\biggr)}=\dfrac{3\Bigl(1+\dfrac t{18}-1-\dfrac{2t}3+o(t)\Bigr)}{2\Bigl(1+\dfrac t{24}-1-t+o(t)\Bigr)}\\ &=\frac{-\dfrac{11t}6+o(t)}{-\dfrac{23t}{24}+o(t)}=\frac{22+o(1)}{23+o(1)}=\frac{22}{23}+o(1) \end{align*} So the limit as $t$ tends to $0$ is equal to $\color{red}{\dfrac{22}{23}}$.

Answer 4

By using the first two terms in the generalized binomial development, $$\frac{\sqrt{t+9}-3 \sqrt[3]{2 t+1}}{\sqrt[3]{t+8}-2 (3 t+1)}= \frac{3\left(\frac t9+1\right)^{1/2}-3 (2 t+1)^{1/3}}{2\left(\frac t8+1\right)^{1/3}-2 (3 t+1)^{1/3}}\approx \frac32\frac{\frac t{18}-\frac{2t}3}{\frac t{24}-t},$$

which tends to $\dfrac{22}{23}$.

Answer 5

Rewrite the limit in the seemingly more complicated way $$ \lim_{t\to0} \frac {\dfrac{\sqrt{t+9}-3}{t}-3\dfrac{\sqrt[3]{2t+1}-1}{t}} {\dfrac{\sqrt[3]{t^3+8}-2}{t}-2\dfrac{\sqrt{2t^3+1}-1}{t}} $$ Each piece in the above expression is the ratio for computing the derivative at zero of a function. If you don't know derivatives yet, you can still rationalize each piece in a simpler way than trying to do it for the expression in the original form.