TITLE: $\lim_{n \to \infty}n(\sqrt[3]{\frac{n+a}{n-1}} - \sqrt[2]{\frac{n+b}{n-1}})$
QUESTION [0 upvotes]: I am self studying calculus and cant solve this question:
$$\lim_{n \to \infty}n(\sqrt[3]{\frac{n+a}{n-1}} - \sqrt[2]{\frac{n+b}{n-1}})$$
I would like to show my work, but I have no idea how to get rid of these roots

REPLY [4 votes]: Let us find $$F(m,p,q)=\lim_{n\to\infty} n\left(\sqrt[m]{\dfrac{n+p}{n+q}}-1\right)$$
Let $\lim_{n\to\infty}\sqrt[m]{\dfrac{n+p}{n+q}}=1$
$\sqrt[m]{\dfrac{n+p}{n+q}}-1=h\implies\dfrac{n+p}{n+q}=(1+h)^m,n=?$
$$F(m,p,q)=\lim_{h\to0}(p-q(1+h)^m)\cdot\lim_{h\to0}\dfrac{h}{(1+h)^m-1}=\cdots=\dfrac{p-q}m $$
Find $$F\left(3,a,-1\right)-F\left(2,b,-1\right)$$

REPLY [1 votes]: You can calculate the limit as follows.
For convenience substitute $n=m+1$:
$$\lim_{n \to \infty}n\left(\sqrt[3]{\frac{n+a}{n-1}} - \sqrt[2]{\frac{n+b}{n-1}}\right) = \lim_{m \to \infty}(m+1)\left(\sqrt[3]{\frac{m+1+a}{m}} - \sqrt[2]{\frac{m+1+b}{m}}\right)$$
$$ = \lim_{m \to \infty}\frac{m+1}{m}\left(\sqrt[3]{m^3+(1+a)m^2} - \sqrt[2]{m^2+(1+b)m}\right)$$
Since, $\lim_{m \to \infty}\frac{m+1}{m} = 1$, you only need to consider $ \sqrt[3]{m^3+(1+a)m^2} - \sqrt[2]{m^2+(1+b)m}$. Substituting $m = \frac 1t$ and considering $t\to 0^+$ reduces the limit to the calculation of two derivatives:
\begin{eqnarray} \sqrt[3]{m^3+(1+a)m^2} - \sqrt[2]{m^2+(1+b)m}
& \stackrel{m =\frac 1t}{=} & \frac{\sqrt[3]{1+(1+a)t} - \sqrt[2]{1+(1+b)t}}{t} \\
& = & \frac{\sqrt[3]{1+(1+a)t} - 1}{t} - \frac{\sqrt[2]{1+(1+b)t} - 1}{t} \\
& \stackrel{t \to 0^+}{\longrightarrow} & \left.\left(\sqrt[3]{1+(1+a)t}\right)' \right|_{t=0} - \left.\left(\sqrt[2]{1+(1+b)t}\right)' \right|_{t=0}  \\
& = & \frac{1+a}{3} - \frac{1+b}{2}
\end{eqnarray}
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