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Find the following limit:

$$\lim_{x\to 1}\frac{\sin{(\pi\sqrt x)}}{\sin{(\pi x)}}$$

My attempt:

$$t:=x-1,\ x \rightarrow 1 \Rightarrow t\rightarrow 0,\ x=t+1$$ $$\lim_{x\to 1}\frac{\sin{(\pi\sqrt x)}}{\sin{(\pi x)}}=\lim_{t\to 0}\frac{\sin{(\pi\sqrt{t+1})}}{\sin{(\pi(t+1))}}=\lim_{t\to 0}\frac{\frac{\sin{(\pi\sqrt{t+1})}}{\pi(\sqrt{t+1})}\cdot \pi \sqrt{(t+1)}}{\frac{\sin{(\pi(t+1))}}{\pi(t+1)}\cdot \pi(t+1)}=\lim_{t\to 0}\frac{1\cdot \pi \sqrt{(t+1)}}{1\cdot\pi(t+1)}=\frac{\pi\sqrt{(0+1)}}{\pi\sqrt{(0+1)}}=1$$

The soulution should be $\frac{1}{2}$. What am I doing wrong?

  • 0
    The limit $\sin(x)/x\to 1$ is for $x\to 0$. – 2017-01-28
  • 0
    Is your question how to compute the limit or where you went wrong or both? – 2017-01-28

4 Answers 4

6

Your mistake is here $$\lim _{ t\to 0 } \frac { \sin { \left( \pi \left( t+1 \right) \right) } }{ \pi \left( t+1 \right) } \neq 1\\ \lim _{ t\to 0 } \frac { \sin { \left( \pi \sqrt { \left( t+1 \right) } \right) } }{ \pi \sqrt { \left( t+1 \right) } } \neq 1$$

4

\begin{eqnarray} \lim_{x\to1}\dfrac{\sin(\pi\sqrt{x})}{\sin(\pi x)}&=&\lim_{x\to1}\dfrac{\sin(\pi (\sqrt{x}-1))}{\sin(\pi(x-1))}=\lim_{x\to1}\dfrac{\sin(\pi(\sqrt{x}-1))}{\pi(\sqrt{x}-1)}\cdot\lim_{x\to1}\dfrac{\pi(\sqrt{x}-1)}{\pi(x-1)}\cdot\lim_{x\to1}\dfrac{\pi(x-1)}{\sin(\pi(x-1))}\\ &=&\lim_{u\to0}\dfrac{\sin u}{u}\cdot\lim_{v\to1}\dfrac{v-1}{v^2-1}\cdot\lim_{w\to0}\dfrac{w}{\sin w}=1\cdot\lim_{v\to1}\dfrac{1}{v+1}\cdot1=\dfrac12. \end{eqnarray}

2

As $\sin(\pi-y)=\sin y$

$$\dfrac{\sin(\pi\sqrt x)}{\sin(\pi x)}=\dfrac{\sin\pi(1-\sqrt x)}{\sin\pi(1-x)}$$

Now set $1-\sqrt x=u$

0

A way to work it out could be using Taylor expansions around $x=1$

Then you get:

\begin{equation} \lim_{x \rightarrow 1}{\dfrac{-\dfrac{1}{2}\pi (x-1) + \dfrac{1}{8}\pi (x-1)^2 + O(x^3)}{-\pi (x-1) + O(x^3)} } \end{equation}

Since the terms with the lower power are the ones that tend slower to 0, they are dominating on the expression; therefore:

\begin{equation} = \lim_{x \rightarrow 1}{\dfrac{-\dfrac{1}{2}\pi (x-1) }{-\pi (x-1)} } = \dfrac{1}{2} \end{equation}