Limits of functions with square root and sin Can someone help me calculate the following limits(without L'Hopital!) :
1) $\lim_{x\to 1 } \frac { \sqrt{x}-1}{\sqrt[3]{x}-1}  $ . I have tried taking the logarithm of this limit, but without any success.
2) $\lim_{x\to\pi} \frac{\sin5x}{\sin3x}$  and the hint is : $\sin(\pi-x)=\sin x , \sin(2\pi-x)=-\sin x$.
Thanks everyone! 
 A: For the first one, set $\sqrt[6]{x} = y$. This means $\sqrt{x} = y^3$ and $\sqrt[3]{x} = y^2$. We then get that
$$\lim_{x \to 1} \dfrac{\sqrt{x}-1}{\sqrt[3]{x}-1} = \lim_{y \to 1} \dfrac{y^3-1}{y^2-1} = \lim_{y \to 1} \dfrac{(y-1)(y^2+y+1)}{(y-1)(y+1)} = \lim_{y \to 1} \dfrac{(y^2+y+1)}{(y+1)} = \dfrac32$$
For the second one, as the hint suggests, let $y = \pi - x$. We then get that
\begin{align}
\lim_{x \to \pi} \dfrac{\sin(5x)}{\sin(3x)} & = \lim_{y \to 0} \dfrac{\sin(5(\pi-y))}{\sin(3(\pi-y))} = \underbrace{\lim_{y \to 0} \dfrac{\sin(5 \pi -5y)}{\sin(3\pi-3y)} = \lim_{y \to 0} \dfrac{\sin(5y)}{\sin(3y)}}_{(\star)}\\
& = \lim_{y \to 0} \dfrac{5 \times \dfrac{\sin(5y)}{5y}}{3 \times \dfrac{\sin(3y)}{3y}} = \dfrac{5 \times \lim_{y \to 0} \dfrac{\sin(5y)}{5y}}{3 \times \lim_{y \to 0} \dfrac{\sin(3y)}{3y}} = \dfrac53\\
\end{align}
where $(\star)$ follows from the fact that $\sin((2n+1)\pi-y) = \sin(y)$ and the last equality comes from the fact that $\displaystyle \lim_{\theta \to 0} \dfrac{\sin(\theta)}{\theta} = 1$.
EDIT
$$\sin((2n+1)\pi-y) = \overbrace{\sin(2n \pi + \pi -y) = \sin(\pi -y)}^{\text{Because $\sin$ is $2 \pi$ periodic.}}$$
Now $$\sin(\pi-y) = \sin(\pi) \cos(y) - \cos(\pi) \sin(y) = 0 - (-1) \sin(y) = \sin(y)$$
A: 1) $\lim_{x\to 1 } \frac { \sqrt{x}-1}{\sqrt[3]{x}-1}=\lim_{x\to 1 } \frac { (\sqrt[]{x}-1)(\sqrt[]{x}+1)}{(\sqrt[3]{x}-1)(\sqrt[]{x}+1)}=\lim_{x\to 1 } \frac { x-1}{(\sqrt[3]{x}-1)(\sqrt[]{x}+1)}= \lim_{x\to 1 } \frac { x^{2/3}+x^{1/3}+1}{(\sqrt[]{x}+1)}=3/2 $ .
2) For the other limit, see Marvis' answer
