Show that $\ x_n$ is correctly defined if $2^{x_n}+3^{x_n}= n$

Question

The sequence $\ (x_n)_{n>=1}$ defined by the relation $2^{x_n}+3^{x_n}= n$ for any $\ n$ bigger or equal to $\ 1$. Show that $\ x_n$ is correctly defined and find the limit of $\dfrac{x_n}{n}$ when the limit of $\ n$ is infinite. I know that $\ x_2$ is $\ 0$ however I feel unable to accomplish the task.

Answer 1

Let $f(t) = 2^t + 3^t$.

Since $f$ is strictly increasing, and the range of $f$ is $(0,\infty)$, it follows that $x_n$ is well defined, as the unique solution to the equation $f(t)=n$.

For the desired the limit, \begin{align*} &2^{x_n}+3^{x_n}=n\\[4pt] \implies\;&2^{x_n} < n\\[4pt] \implies\;&x_n < \log_2(n)\\[4pt] \implies\;&\frac{x_n}{n} < \frac{\log_2(n)}{n}\\[4pt] \end{align*} Also, if $n > 2$, the equation $2^{x_n}+3^{x_n}=n$ implies $x_n > 0$.

Hence, for $n > 2$, we have $$0 < \frac{x_n}{n} < \frac{\log_2(n)}{n}$$ It follows that $$\lim_{n\to \infty} \frac{x_n}{n} = 0$$

Answer 2

The first part of the question simply asks

Show that, for every integer $n \geqslant 1$, the equation $2^x+3^x=n$ has exactly one solution. Call it $x_n$.

Indeed, if there was no solution, $x_n$ could not be assigned any value, while if there was more than a solution, then there would be ambiguity on the value of $x_n$.

To show that $x_n$ is well-defined, you can say that $f(x)=2^x+3^x$ is a one-to-one function (as the sum of two increasing functions, it is increasing, hence one-to-one), so the equation $2^x+3^x=n$ can have at most one solution. The range of $f$ is $(0,\infty)$, so the equation has a solution for each positive integer $n$.

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For the second part, we can make a rough estimate that $x_n<\log_2(n)$. Indeed, we have $$2^{x_n}<2^{x_n}+3^{x_n}=n$$

Therefore, $0<\frac{x_n}{n}<\frac{\log_2(n)}{n}$ for $n$ large enough. Since you probably know that $\lim_{n\to\infty}\frac{\log_2(n)}{n}=0$ (using L'Hospital's Rule for example), we obtain by the Squeeze Theorem that $\lim_{n\to\infty}\frac{x_n}{n}=0$.