Computing: $\lim_{x\to\infty}\frac{\sqrt{1-\cos^2\frac{1}{x}}\left(3^\frac{1}{x}-5^\frac{-1}{x}\right)}{\log_2(1+x^{-2}+x^{-3})}$

Question

Find the following limit:

$$\lim_{x\to\infty}\frac{\sqrt{1-\cos^2\frac{1}{x}}\left(3^\frac{1}{x}-5^\frac{-1}{x}\right)}{\log_2(1+x^{-2}+x^{-3})}$$

I'm not sure whether my solution is correct.

$t:=\frac{1}{x}$

$$\lim_{x\to\infty}\frac{\sqrt{1-\cos^2\frac{1}{x}}\left(3^\frac{1}{x}-5^\frac{-1}{x}\right)}{\log_2(1+x^{-2}+x^{-3})}=\lim_{t\to 0}\frac{\sqrt{1-\cos^2 t}\left(3^t-5^{-t}\right)}{\log_2(1+t^2+t^3)}$$

$$=\lim_{t\to 0}\frac{\frac{\sqrt{1-\cos^2t}}{\sqrt t^2}\cdot t\cdot\left(\frac{3^t-1}{t}\cdot t+(-t)\frac{(-5)^{-t}+1}{-t}\right)}{\log_2(1+t^2+t^3)}$$

$$=\frac{1}{2}(\ln 3+\ln 5)\left[\lim_{t\to 0}\log_2(1+t^2+t^3)^\frac{1}{t^2}\right]^{-1}=\frac{1}{2}(\ln3+\ln 5)\left(e^{{\lim_{t\to 0}\frac{t^2+t^3}{t^2}}^{-1}}\right)^{-1}=\frac{\ln3+\ln5}{2e}$$

Answer 1

Your idea is very good; the limit you get is for $t\to0^+$, so $\sqrt{1-\cos^2t}=\sin t$ and you get $$ \lim_{t\to0^+}\frac{\sin t(3^t-5^{-t})}{\log_2(1+t^2+t^3)}= \lim_{t\to0^+}\frac{\sin t}{t}\frac{3^t-5^{-t}}{t}\frac{t^2\log 2}{\log(1+t^2+t^3)} $$ (where “log” denotes the natural logarithm) and you can compute separately the limit of the three factors. The first is known to be $1$. Then $$ \lim_{t\to0^+}\frac{3^t-5^{-t}}{t}=\log 3+\log 5 $$ because it's the derivative at $0$ of $f(t)=3^t-5^{-t}$. Alternatively, write it as $$ \lim_{t\to0^+}\left(\frac{3^t-1}{t}+\frac{5^t-1}{t}\frac{1}{5^t}\right) $$ and use the fundamental limits (which is basically what you did).

For the last one, apply l’Hôpital (or Taylor): $$ \lim_{t\to0^+}\frac{2t\log2}{\dfrac{2t+3t^2}{1+t^2+t^3}}= \lim_{t\to0^+}\frac{2(1+t^2+t^3)\log2}{2+3t}=\log2 $$ So finally you get $(\log3+\log5)\log2=(\log 15)(\log 2)$

Answer 2

$\begin{array}\\ \dfrac{\sqrt{1-\cos^2\frac{1}{x}} (3^\frac{1}{x}-5^\frac{-1}{x})}{\log_2(1+x^{-2}+x^{-3})} &=\dfrac{\sin(1/x) (e^\frac{\ln 3}{x}-e^\frac{-\ln 5}{x})}{(1/\ln 2)\ln(1+x^{-2}+x^{-3})}\\ &=\dfrac{(1/x+O(1/x^3)) ((1+\ln 3/x+O(1/x^2)-(1-\ln 5/x+O(1/x^2))}{(1/\ln 2)(x^{-2}+x^{-3}+O(x^{-4})}\\ &=\dfrac{(1/x+O(1/x^3)) ((\ln 3+\ln 5)/x+O(1/x^2))\ln 2}{x^{-2}(1+O(x^{-3}))}\\ &=\dfrac{1+O(1/x)) ((\ln 3+\ln 5)+O(1/x))\ln 2}{1+O(1/x)}\\ &=(\ln 2(\ln 15)+O(1/x))(1+O(1/x))\\ &=(\ln 2)(\ln 15)+O(1/x)\\ \end{array} $