Finding: $\lim\limits_{x\to 0} \bigg(\frac{1-\sqrt[5]{(1+x)^3}}{x}+(1+3\sin(x))^{\frac{2}{\sin(x)}}\bigg)$ I'm quite confused by this example of finding the limit$$\lim\limits_{x\to 0}\frac{1-\sqrt[5]{(1+x)^3}}{x}+(1+3\sin(x))^{\frac{2}{\sin(x)}}$$ 
I know that if the limit goes to 0, I need to multiply by the conjugate. 
I know to do it in term of $x^{1/2}$, but how I do it in this case?

What is the derivative of $g:x\mapsto (1+x)^{3/5}$ at $x=0$?  We have, 
$$ \lim_{x\to0} \frac{1-(1+x)^{3/5}}{x}= \lim_{x\to0} \frac{(1+x)^{3/5}-1}{x-0}=-g'(0)= -\frac35$$
on the other hand, since $t=\sin x\to 0$ as $x\to 0$   we have,
 $$\lim_{x\to0}(1+3\sin(x))^{\frac{2}{\sin(x)}} =\lim_{t\to0} (1+3t)^{\frac{2}{t}} =\exp\left(6\frac{\ln(1+3t)}{3t}\right)=e^6$$
