# 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!

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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 $$\sin(\pi-y) = \sin(\pi) \cos(y) - \cos(\pi) \sin(y) = 0 - (-1) \sin(y) = \sin(y)$$

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Can you please explain to me how you went from $5\pi-5y$ to $5y$ ? Thanks ! – joshua Nov 25 '12 at 8:33
The numerator of the first limit has $sqrt(x)-1$ – Amr Nov 25 '12 at 8:38
@Amr Yes. I realized it and I was editing it when you commented. Thanks. – user17762 Nov 25 '12 at 8:39

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

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