Find the limit $\lim_{x \to 0} (2^x + \sin (3x)) ^{\cot(3x)}$ Please help, I have already tried every thing I can, but nothing works. I have no I idea what to do.
$$\lim_{x \to 0} \; (2^x + \sin (3x)) ^{\cot(3x)}$$
 A: $\because \sin{3x} \approx 3x, \cos{3x} \approx 1$, as $x \to 0$
$\cot{3x} \approx \frac{1}{3x}$
$\therefore \lim \limits_{x \to 0} \,(2^x+\sin{3x})^{\cot{3x}}$
$\approx \lim \limits_{x \to 0} \,(2^x+3x)^\frac{1}{3x}$
$=\lim \limits_{x \to 0} \,2^\frac{1}{3}(1+\frac{3x}{2^x})^{\frac{2^x}{3x}{\frac{1}{2^x}}}$
$= 2^\frac{1}{3} e$
A: we will use the fact that $$ (1 + small)^{BIG} = e^{small* BIG} + \ldots $$ which follows from the definition of $e.$ 
first, $$\begin{align}2^x &= e^{xln2} = 1 + x\ln 2 + \ldots\\
\sin 3x  &= 3x + \ldots\\
\cot (3x) &= \frac 1{3x} + \ldots  
\end{align}$$
therefore $$ (2^x+ \sin 3x)^{\cot 3x} =\left(1 + x(\ln 2 + 3)+\ldots\right)^{\frac 1{3x}+\ldots} = e^{(3+\ln 2)/3} + \ldots = 2^{1/3}e+ \ldots$$ 
and the required limit  $\lim_{x \to 0} \; (2^x + \sin (3x)) ^{\cot(3x)} = 2^{1/3}e.$
A: Find the limit of the logarithm first, i. e. of
\begin{align*}
\cot(3x)\ln(2^x&+\sin 3x)=\cot 3x\Bigl(x\ln2+\ln\frac{\sin3x}{2^x}\Bigr)\\
&=\frac{\cos3x}{\sin3x}\biggl(x\ln2+\ln\Bigl(1+\frac{\sin3x}{2^x}\Bigr)\biggr)=\frac{\cos3x}{\sin3x}\biggl(x\ln2+\frac{\sin3x}{2^x}+o\Bigl(\frac{\sin3x}{2^x}\Bigr)\biggr)\\
&=\cos3x\biggl(\dfrac{x\ln2}{\sin 3x}+\frac1{2^x}+o\Bigl(\frac1{2^x}\Bigr)\biggr)\to\frac{\ln2}{3}+1,
\end{align*}
hence
$$\lim_{x \to 0} \; (2^x + \sin (3x)) ^{\cot(3x)}=\sqrt[3]2\mkern1.5mu\mathrm e.$$
A: Rewritting the limit using the exponential function $e^x$ using the logarthmic identity $e^{\ln x} = x$.
\begin{align}
\lim_{x \to 0} (2^x + \sin(3x))^{\cot(3x)} &= \lim_{x \to 0} \exp(\cot(3x)\ln(2^x + \sin(3x)))\\
&= \exp( \lim_{x \to 0} \cot(3x)\ln(2^x + \sin(3x)))\\
&= \exp(\lim_{x \to 0} \frac{\ln(2^x + \sin(3x))}{\tan(3x)} )\\
\end{align}
Now apply L'Hopital's rule
\begin{align}
\lim_{x \to 0} \frac{\ln(2^x + \sin(3x))}{\tan(3x)} &= \lim_{x \to 0} \frac{\frac{3\cos(3x) + 2^x \ln(2)}{\sin(3x) + 2^x}}{3\sec^2{3x}}\\
&= \frac{3\cos(0) + 2^0 \ln(2)}{(\sin(0) + 2^0)(3\sec^2{0})} =\frac{3 + \ln(2)}{3}
\end{align}
Plugging this in yields $\lim_{x \to 0} (2^x + \sin(3x))^{\cot(3x)} = e^{\frac{3 + \ln(2)}{3}} = e^1e^\frac{\ln(2)}{3} = 2^\frac{1}{3}e = \sqrt[3]{2}e$
A: While evaluating limit of expressions of type $\{f(x)\}^{g(x)}$ it is best to take logarithms. Let $L$ be the desired limit so that
\begin{align}
\log L &= \log\left(\lim_{x \to 0}(2^{x} + \sin 3x)^{\cot 3x}\right)\notag\\
&= \lim_{x \to 0}\log(2^{x} + \sin 3x)^{\cot 3x}\text{ (via continuity of log)}\notag\\
&= \lim_{x \to 0}\cot 3x\log(2^{x} + \sin 3x)\notag\\
&= \lim_{x \to 0}\frac{\log(2^{x} + \sin 3x)}{\tan 3x}\notag\\
&= \lim_{x \to 0}\frac{\log(2^{x} + \sin 3x)}{3x}\cdot\frac{3x}{\tan 3x}\notag\\
&= \frac{1}{3}\lim_{x \to 0}\frac{\log(1 + 2^{x} - 1 + \sin 3x)}{x}\notag\\
&= \frac{1}{3}\lim_{x \to 0}\frac{\log(1 + 2^{x} - 1 + \sin 3x)}{2^{x} - 1 + \sin 3x}\cdot\frac{2^{x} - 1 + \sin 3x}{x}\notag\\
&= \frac{1}{3}\lim_{x \to 0}1\cdot\left(\frac{2^{x} - 1}{x} + 3\cdot\frac{\sin 3x}{3x}\right)\notag\\
&= \frac{1}{3}(\log 2 + 3) = 1 + \frac{\log 2}{3}\notag
\end{align}
and hence $L = 2^{1/3}e$.
