How to solve this limit without L'Hopital? $\lim_{n\to \infty } (6^{1/n}+4^{1/n}-3^{1/n})^n$ I assume it would work since I think the three components exists in their own but I'm not really sure if it works like that but this is the method I tried, can anyone confirm that I didn't mess up?
$$A=\lim_{n\to \infty }  (6^{1/n}+4^{1/n}-3^{1/n})^n$$
$$\log (A)=\log (\lim_{n\to \infty }  (6^{1/n}+4^{1/n}-3^{1/n}))^n$$
$$\log (A)= n\cdot \log (\lim_{n\to \infty }  (6^{1/n}+4^{1/n}-3^{1/n}))$$
$$\log(A)= n\cdot \log( \lim_{n\to \infty }6^{1/n}+\lim_{n\to \infty } 4^{1/n}- \lim_{n\to \infty }3^{1/n}))$$
Is this allowed or am I messing up somewhere?
 A: \begin{align}
\lim_{n\to\infty}(6^{1/n}+4^{1/n}-3^{1/n})^n&=\lim_{n\to\infty}(e^{\ln6/n}+e^{\ln4/n}-e^{\ln3/n})^n\\
&=\lim_{n\to\infty}\left(\left(1+\frac{\ln6}n\right)^{n/n}+\left(1+\frac{\ln4}n\right)^{n/n}-\left(1+\frac{\ln3}n\right)^{n/n}\right)^n\\
&=\lim_{n\to\infty}\left(\left(1+\frac{\ln6}n\right)+\left(1+\frac{\ln4}n\right)-\left(1+\frac{\ln3}n\right)\right)^n\\
&=\lim_{n\to\infty}\left(1+\frac{\ln8}n\right)^n\\
&=e^{\ln8}\\
&=8
\end{align}
A: Not allowed you have to take n to infinity at the same time. To solve the limit try using the Sandwich Theorem instead.
A: $A=\lim_{n\to\infty}6(1+(\dfrac{2}{3})^{1/n}-(\dfrac{1}{2})^{1/n})^n=6e^{\lim_{n\to\infty}n\left[(\dfrac{2}{3})^{1/n}-(\dfrac{1}{2})^{1/n}\right]}$
Now $\dfrac{n(4^{1/n}-3^{1/n})}{6^{1/n}}$ is the quantity whose limit we are interested in.
Consider the function $f(x)=(2/3)^x$ then $f$ is differentiable at $0$ and we have, $\lim_{h\to0}\dfrac{f(h)-f(0)}{h}=f'(0)=\log(2/3)$ i.e. $\lim_{h\to0}\dfrac{(2/3)^h-1}{h}=\log(2/3)$.
Similarly consider the function $g(x)=(1/2)^x$ and analogously we have $\log(1/2)=g'(0)=\lim_{h\to0}\dfrac{(1/2)^h-1}{h}$.
Hence we have the result $\lim_{h\to0}\dfrac{f(h)-g(h)}{h}=\lim_{h\to0}\dfrac{f(h)-f(0)}{h}-\lim_{h\to0}\dfrac{g(h)-g(0)}{h}=\log(2/3)-\log(1/2)=\log(4/3)$
So $\dfrac{n(4^{1/n}-3^{1/n})}{6^{1/n}}=\dfrac{(2/3)^{1/n}-(1/2)^{1/n}}{1/n}\to\log\dfrac{4}{3}$ (here $h=1/n\to0$).
So the final limit is $6e^{\log4/3}=8$.
A: If $L$ is the desired limit then we can proceed as follows
\begin{align}
\log L &= \log\left\{\lim_{n \to \infty}(6^{1/n} + 4^{1/n} - 3^{1/n})^{n}\right\}\notag\\
&= \lim_{n \to \infty}\log(6^{1/n} + 4^{1/n} - 3^{1/n})^{n}\text{ (via continuity of log)}\notag\\
&= \lim_{n \to \infty}n\frac{\log(1 + 6^{1/n} + 4^{1/n} - 3^{1/n} - 1)}{6^{1/n} + 4^{1/n} - 3^{1/n} - 1}\cdot(6^{1/n} + 4^{1/n} - 3^{1/n} - 1)\notag\\
&= \lim_{n \to \infty}n(6^{1/n} + 4^{1/n} - 3^{1/n} - 1)\notag\\
&= \lim_{n \to \infty}n(6^{1/n} - 1) + n(4^{1/n} - 1) - n(3^{1/n} - 1)\notag\\
&= \log 6 + \log 4 - \log 3\notag\\
&= \log 8\notag
\end{align}
Hence $L = 8$. We have used the following standard limits $$\lim_{x \to 0}\frac{\log(1 + x)}{x} = 1,\lim_{n \to \infty}n(x^{1/n} - 1) = \log x$$
