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.
 A: $$\lim_{x\to \infty}\left(4\times6^x-3\times10^x+8\times15^x\right)^{1/x}=\lim_{x\to \infty}(15^x)^{1/x}\left(4\times\frac{6^x}{15^x}-3\times \frac{10^x}{15^x}+8\right)^{1/x}$$
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.
A: 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.
A: The key is that in the equation $4*6^x-3*10^x+8*15^x$, the $8*15^x$ part will tend to be $100$% of the entire equation.  The reason is that $\frac{4*6^x}{8*15^x} \to 0$ as $x \to \infty$ and $\frac{-3*10^x}{8*15^x} \to 0$ as $x \to \infty$ because the number being raised to the power is smaller.
So that leaves us with $\lim_{x\to\infty}(8*15^x)^{1/x}$.
We can break this up into $\lim_{x\to\infty}(8^{1/x}15)$ because $(15^x)^{1/x} = 15$.
And because $8^{1/x} \to 1$ that just leaves us with $\lim_{x\to\infty}(15)$ which is 15.
If you have any questions just ask! :)
