Find the value of this Limit Find the value of,
$$\lim_{x\to \infty}\left(\dfrac{x+15}{x+3}\right)^{\large{\ln\left(\sin{\frac{1}{x}}\right)}}$$
I've tried putting it in exponential log form and then apply l'Hospital rule but that derivative is giving me a headache.
 A: The obvious substitution to make is $x=1/t$, that transforms your limit into
$$
\lim_{t\to0^+}\left(\frac{15+1/t}{3+1/t}\right)^{\ln\sin t}
$$
Now it's time to try the logarithm of this:
$$
\lim_{t\to0^+}\ln\sin t\ln\left(\frac{15+1/t}{3+1/t}\right)=
\lim_{t\to0^+}\ln\sin t\ln\left(\frac{15t+1}{3t+1}\right)
$$
Consider, for the moment,
$$
\lim_{t\to0^+}\frac{1}{t}\ln\left(\frac{15t+1}{3t+1}\right)=
\lim_{t\to0^+}\frac{\ln(15t+1)-\ln(3t+1)}{t}
$$
which is good for using l'Hôpital's theorem:
$$
\lim_{t\to0^+}\frac{\ln(15t+1)-\ln(3t+1)}{t}\overset{\mathrm{(H)}}{=}
\lim_{t\to0^+}\frac{\dfrac{15}{15t+1}-\dfrac{3}{3t+1}}{1}=12
$$
On the other hand,
$$
\lim_{t\to0^+}t\ln\sin t=\lim_{t\to0^+}\frac{\ln\sin t}{1/t}
\overset{\mathrm{(H)}}{=}
\lim_{t\to0^+}\frac{\dfrac{\cos t}{\sin t}}{-\dfrac{1}{t^2}}
=
\lim_{t\to0^+}-t\cos t\frac{t}{\sin t}=0
$$
Thus
$$
\lim_{t\to0^+}\ln\sin t\ln\left(\frac{15t+1}{3t+1}\right)
=
\lim_{t\to0^+}(t\ln\sin t)\cdot\left(\frac{1}{t}\ln\left(\frac{15t+1}{3t+1}\right)\right)=0\cdot 12=0
$$
Therefore your original limit is $\boxed{e^0=1}$

Note
Why multiplying and dividing by $t$? The reason is that
$$
\ln\frac{15t+1}{3t+1}=\ln(15t+1)-\ln(3t+1)=15t+o(t)-(3t+o(t))=12t+o(t)
$$
so “it counts as $12t$”. Dividing it by $t$ leaves a finite nonzero limit, which is never a cause of indetermination.
Thus we are left with $\lim_{t\to0^+}t\ln\sin t$ and this is very similar to $\lim_{t\to0^+}t\ln t=0$. Without l'Hôpital's theorem one could do by
$$
\lim_{t\to0^+}t\ln\sin t=
\lim_{t\to0^+}t\ln\left(t\frac{\sin t}{t}\right)=
\lim_{t\to0^+}\left(t\ln t+ t\ln\frac{\sin t}{t}\right)
$$
and both summands have zero limit.
A: Take $y = \frac{1}{x}$ then $y \to 0$ when $x \to \infty$ and 
$$\lim_{x\to \infty} \Bigg(\frac{1 + \frac{15}{x}}{1+\frac{3}{x}}\Bigg)^{\ln(\sin \frac{1}{x})} =\lim_{y\to 0} \Bigg(\frac{1 + 15y}{1+3y}\Bigg)^{\ln(\sin y)} $$
A: It's enough to determine the limit of the logarithm: $$\ln\Bigl(\sin\dfrac{1}{x}\Bigr)\ln\frac{x+15}{x+3}=\ln(\sin y)\bigl(\ln(1+15y)-\ln(1+3y)\bigr)\underset 0 \sim 12y\ln(\sin y)$$
if we set $y=\dfrac 1 x$.
Now set $u=\sin y, u\in [-\frac\pi 2,\frac\pi 2]$, i.e. $y=\arcsin u$. Then
$$12y\ln(\sin y)=12\arcsin u\ln u\underset{0_+} \sim 12u\ln u\underset{u\to 0_+}\longrightarrow 0_{-}.$$
Hence the original expression tends to $1$ from below.
