Ceiling function and limit problem $\lim_{n \to \infty} \frac {\lceil 4^{n+\frac{\log \frac 83}{\log4}} \rceil}{\lceil 4^{n+0.707519} \rceil}$

Question

I want to calculate this limit:

$$\lim_{n \to \infty} \frac {\lceil 4^{n+\frac{\log \frac 83}{\log4}} \rceil}{\lceil 4^{n+\lambda} \rceil}$$

where, $\lambda$ is a constant and I know $6$ digits. $\lambda =0.707519$

Is it possible to find this limit for $\lambda=0.797519?$ Is this limit equals to $1$?

$$\lim_{n \to \infty} \frac {\lceil 4^{n+\frac{\log \frac 83}{\log4}} \rceil}{\lceil 4^{n+0.707519} \rceil}$$

$\lambda =0.707519$, can we say that, this limit equal to $1$? I tried WolframAlpha but, the Wolfram can not find..I have no Method how can I can calculate this limit. I am stuck

Answer 1

As $n\to\infty$, the relative error induced by dropping the ceiling function goes to $0$. We conclude $$\lim_{n\to\infty}\frac{\left\lceil4^{n+\frac{\log\frac83}{\log 4}}\right\rceil}{\left\lceil4^{n+\lambda}\right\rceil}=\lim_{n\to\infty}\frac{4^{n+\frac{\log\frac83}{\log 4}}}{4^{n+\lambda}}=4^{\frac{\log\frac83}{\log 4}-\lambda.}$$ Hence the limit is $1$ iff $\lambda=\frac{\log\frac83}{\log 4}\approx 0.707518749639421909273130528026$