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.


3 Answers 3


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

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! :)

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

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .