Computing: $\lim_{x\to\infty}\frac{\sqrt{1-\cos^2\frac{1}{x}}\left(3^\frac{1}{x}-5^\frac{-1}{x}\right)}{\log_2(1+x^{-2}+x^{-3})}$ 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.
$t:=\frac{1}{x}$
$$\lim_{x\to\infty}\frac{\sqrt{1-\cos^2\frac{1}{x}}\left(3^\frac{1}{x}-5^\frac{-1}{x}\right)}{\log_2(1+x^{-2}+x^{-3})}=\lim_{t\to 0}\frac{\sqrt{1-\cos^2 t}\left(3^t-5^{-t}\right)}{\log_2(1+t^2+t^3)}$$
$$=\lim_{t\to 0}\frac{\frac{\sqrt{1-\cos^2t}}{\sqrt t^2}\cdot t\cdot\left(\frac{3^t-1}{t}\cdot t+(-t)\frac{(-5)^{-t}+1}{-t}\right)}{\log_2(1+t^2+t^3)}$$
$$=\frac{1}{2}(\ln 3+\ln 5)\left[\lim_{t\to 0}\log_2(1+t^2+t^3)^\frac{1}{t^2}\right]^{-1}=\frac{1}{2}(\ln3+\ln 5)\left(e^{{\lim_{t\to 0}\frac{t^2+t^3}{t^2}}^{-1}}\right)^{-1}=\frac{\ln3+\ln5}{2e}$$ 
 A: 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 $(\log3+\log5)\log2=(\log 15)(\log 2)$
A: $\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}
$
A: Fast solution:
Using the minimum number of terms of the Taylor developments,
$$\lim_{t\to 0^+}\frac{\sin t\,(e^{t\log 3}-e^{-{t\log 5}})}{\log_2(1+t^2+t^3)}$$ has


*

*at the numerator a factor $t$, and the difference between $1+t \log3$ and $1-t\log 5$;

*at the denominator, $\dfrac{t^2}{\log2}$.
Hence the limit is 
$$\log2\log15.$$

Without Taylor, 
we can replace $\sin t$ by $t$ (thanks to the limit of $\dfrac{\sin t}t$), and $e^{t\log 3}-e^{-t\log 5}$ by $e^{t\log 15}-1$ by pulling out a factor $e^{-5t\log 5}$. Then $e^{t\log 15}-1=\dfrac{e^{t\log 15}-1}{t\log 15}t\log 15$ can be replaced by $t\log 15$.
Then $\lim_{t\to0}\log(1+t^2+t^3)=\lim_{t\to0}\dfrac{\log(1+t^2+t^3)}{t^2+t^3}(t^2+t^3)$.
Finally, we have
$$\log2\log15\lim_{t\to0}\frac{t^2}{t^2+t^3}.$$
A: Here is a solution without using Taylor series or applying L'Hôpital's rule:
$$\lim_{x\to\infty}\frac{\sqrt{1-\cos^{2}\left(\frac{1}{x}\right)}\left(3^{\frac{1}{x}}-5^{\frac{-1}{x}}\right)}{\log_{2}\left(1+x^{-2}+x^{-3}\right)}$$$$=\lim_{x\to\infty}\frac{\sin\left(\frac{1}{x}\right)}{\color{blue}{\frac{1}{x}}}\cdot\frac{\color{blue} {x^{-2}+x^{-3}}}{\log_{2}\left(1+x^{-2}+x^{-3}\right)}\cdot\left(\frac{3^{\frac{1}{x}}-1}{\color{blue} {\frac{1}{x}}}+\frac{5^{\frac{-1}{x}}-1}{\color{blue}{\frac{-1}{x}}}\right)\cdot\frac{1}{\color{blue}{x^{-1}+1}}$$$$=\ln\left(2\right)\ln\left(15\right)$$
