Limit of sum of exponential functions under root

$$\lim_{x\to \infty}\left(4\times6^x-3\times10^x+8\times15^x\right)^{1/x}=\lim_{x\to \infty}(15^x)^{1/x}\left(4\times\frac{6^x}{15^x}-3\times \frac{10^x}{15^x}+8\right)^{1/x}$$

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.

Write $(4∗6^x−3∗10^x+8∗15^x)^{1/x}$ as $15\cdot (4\cdot (\frac{6}{15}) ^ x - 3 \cdot (\frac{10}{15})^x + 8)^{1/x}$ .

Observe that each term inside main brackets except $8$ goes to $0$ as $x \to \infty$ and $8^0 = 1$.

So the limit value is $15$.

If you are not satisfied with the method, use binomial theorem to solve the problem in more rigorous way.