Find the limit of the sequence of functions: $f_n (x) = \frac{\ln (2^n +x^n)}{n}, x \ge 0$

Question

So, the given function: $f_n (x) = \frac{\ln(2^n +x^n)}{n}, x \ge 0$.

For $|x| < 1$ the limit would be: $\lim_{n \to \infty} \frac{\ln(2^n +x^n)}{n} = \lim_{n \to \infty}\frac{\ln (2^n + 0)}{n} = \frac{\infty}{\infty}$. Can I apply L'Hospital rule here? Then it would be: $\lim_{n \to \infty} \frac{e^{\ln(2) n} \cdot \ln(2)}{2^n} = \lim_{n \to \infty}\frac{\ln(2)^{n+1}}{2^n}$ It already looks nightmerish, so I assume that the method is not the best one in this case. What would be the best way to deal with this limit problem?

Answer 1

$$ \frac{\log(2^n+x^n)}{n}=\log((2^n+x^n)^{1/n}) $$If $0<x<2$, we have $$ \log((2^n+x^n)^{1/n}) = \log(2(1+(x/2)^n)^{1/n}) = \log(2) + \log((1+(x/2)^n)^{1/n}) $$The inner term approaches $1$ by LHR. Similarly, if $x>2$, you can factor out $x$ from each term and conclude that the limit is $\log(x)$.

Answer 2

I would have distinguish 3 cases :

First $|x|<2$

Let's write

$$\dfrac{\ln(2^n+x^n)}{n}=\dfrac{\ln(2^n)+\ln(1+(\dfrac{x}{2})^n)}{n}$$

Frome here you can apply asymptotic of the logarithm development because $\dfrac{x^n}{2^n}$ tends to zero.

Second $x>2$ (Note that there can't have negative values here for $x$ for the log to be defined.

$$\dfrac{\ln(2^n+x^n)}{n}=\dfrac{\ln(x^n)+\ln(1+(\dfrac{2}{x})^n)}{n} $$

And same here, apply asymptotic development.

Third $x=2$

$$ \dfrac{\ln(2^n+x^n)}{n}=\dfrac{(n+1)\ln(2)}{n} $$

And you know how to end it from here.

Answer 3

From $\frac{1}{1+t}\leq \log(1+t)\leq t$ for all $t>-1$, we have that

$$n\log(2)+\frac{(x/2)^n}{1+(x/2)^n}\leq\log(2^n+x^n)=n\log(2)+\log(1+x^n2^{-n})\leq n\log(2)+\frac{x^n}{2^n}$$ and $$n\log(x)+\frac{(2/x)^n}{1+(2/x)^n}\leq \log(2^n+x^n)=n\log(x)+\log(1+2^nx^{-n})\leq n\log(x)+\frac{2^n}{x^n}$$

Thus, for $0\leq x\leq 2$ $$0 \leq \frac1n\log(2^n+x^n)-\log2\leq\frac1n$$ and for $2<x<\infty$ $$0 \leq \frac1n\log(2^n+x^n)-\log x\leq \frac1n$$

Thus $$\frac1n\log(2^n+x^n)\xrightarrow{n\rightarrow\infty}\log(\max(2,x))$$ uniformly on $[0,\infty)$.

Answer 4

For $x\ge 2$ we have $$\log x< {1\over n}\log(2^n+x^n)\\ \le {1\over n}\log(2x^n)={\log 2\over n}+\log x$$ Similarly for $0\le x<2$ we get $$\log 2\le {1\over n}\log(2^n+x^n)\\ \le {1\over n}\log(2\cdot 2^n)={\log 2\over n}+\log 2$$ Hence for $f(x)=\max (\log x,\log 2)$ we obtain $$0\le f_n(x)-f(x)\le {\log 2\over n}$$ Thus the convergence is uniform on $[0,\infty).$