Limit of sum of exponential functions under root How to solve this limit?
$$\underset{x\to \infty }{\text{lim}}\left(4*6^x-3*10^x+8*15^x\right)^{1/x}$$
It is equal $15$ and it seems obvious that it is so. I just can not write it mathematically.
I tried to get rid of $1/x$ in exponent:
$$\underset{x\to \infty }{\text{lim}}\left(4*6^x-3*10^x+8*15^x\right)^{1/x}=\exp \left(\underset{x\to \infty }{\text{lim}}\frac{\log \left(4*6^x-3*10^x+8*15^x\right)}{x}\right)$$
Then applied L'Hôpital's rule:
$$\frac{\partial \log \left(4*6^x-3*10^x+8*15^x\right)}{\partial x}=\frac{4*6^x (\log 6)+8*15^x (\log 15)-3*10^x (\log 10)}{4*6^x-3*10^x+8*15^x}$$
So we have:
$$\underset{x\to \infty }{\text{lim}}\left(4*6^x-3*10^x+8*15^x\right)^{1/x}=\\\exp \left(\underset{x\to \infty }{\text{lim}}\frac{4*6^x \log (6)-3*10^x \log (10)+8*15^x \log (15)}{4*6^x-3*10^x+8*15^x}\right)$$
I can apply the rule again but it only gets more complicated.
I was thinking also about some substitution but can not figure out what substitution to use.

$$\lim_{x\to \infty}\left(4\times6^x-3\times10^x+8\times15^x\right)^{1/x}=\lim_{x\to \infty}(15^x)^{1/x}\left(4\times\frac{6^x}{15^x}-3\times \frac{10^x}{15^x}+8\right)^{1/x}$$
and we have $$(15^x)^{1/x}\left(4\times\frac{6^x}{15^x}-3\times \frac{10^x}{15^x}+8\right)^{1/x}<15(4\times 1-3\times 0+8)^{1/x}\to 15$$
$$(15^x)^{1/x}\left(4\times\frac{6^x}{15^x}-3\times \frac{10^x}{15^x}+8\right)^{1/x}>15(4\times 0-3\times 1+8)^{1/x}\to 15$$
Now apply squeeze theorem.
