# 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.

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.

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! :)

• I was thinking similarly as you and thought it was obvious it must be $15$, but did not consider it to be rigorous enough. Commented Sep 25, 2020 at 17:24