# Limit using L'Hopital's Rule

Find $$\lim_{x \to 1} \sqrt{x-1}^{\,\sin(πx)}$$ using L'Hopital's Rule. Initially I get $0^0$ so I know I need to use the rule, but I don't know where to begin. Could you help me out with some steps on how to solve this limit?

Let $u(x)=\sqrt{x-1}^{\sin(\pi x)}$. We have $$\begin{array}{lll} \lim_{x\rightarrow 1^+}\ln u(x)&=&\lim_{x\rightarrow 1^+}\sin(\pi x)\ln(\sqrt{x-1})\\ &=&\frac{1}{2}\lim_{x\rightarrow 1^+}\frac{\ln(x-1)}{\frac{1}{\sin(\pi x)}}\\ &\overset{H^\prime}{=}&\frac{1}{2}\lim_{x\rightarrow 1^+}\frac{\frac{1}{x-1}}{\frac{-\pi\cos(\pi x)}{\sin^2(\pi x)}}\\ &=&\frac{1}{2}\lim_{x\rightarrow 1^+}\frac{-\sin^2(\pi x)}{\pi(x-1)\cos(\pi x)}\\ &=&\frac{1}{2}\lim_{x\rightarrow 1^+}\frac{-\sin^2[\pi (x-1)]}{\pi(x-1)\cos(\pi x)}\\ &=&\frac{1}{2}\lim_{x\rightarrow 1^+}\frac{-\pi^2 (x-1)^2}{\pi(x-1)\cos(\pi x)}\\ &=&0. \end{array}$$ Hence $\displaystyle\lim_{x\rightarrow 1^+}\sqrt{x-1}^{\sin(\pi x)}=\lim_{x\rightarrow 1^+}u(x)=\lim_{x\rightarrow 1^+}e^{\ln u(x)}=e^{\displaystyle\lim_{x\rightarrow 1^+}\ln u(x)}=e^0=1.$
Let $L =\lim_{x \to 1} \sqrt{x-1}^{\sin(πx)}$. Now take Log on both sides and try now using L'hopitals rule. At the you have to solve for L.
I assume you meant $$\lim_{x \to 1^+} \sqrt{x-1}^{\sin(\pi x)}$$ You could get away without L'hopitals rule. $$\underbrace{\lim_{x \to 1^+} \sqrt{x-1}^{\sin(\pi x)} = \lim_{t \to 0} t^{\sin(\pi(1+t^2))}}_{x = 1 + t^2} = \lim_{t \to 0} t^{\sin(\pi t^2)}$$ $$f(t) = t^{\sin(\pi t^2)} \implies \log(f(t)) = \sin \left(\pi t^2 \right) \log(t)$$ $$\left \vert \sin \left(\pi t^2 \right) \log(t) \right \vert \leq \left \vert \pi t^2 \log(t) \right \vert$$ Now $$\displaystyle \lim_{t \to 0} t^2 \log(t) = 0 \text{ (Why?)}$$ Hence, $$\lim_{t \to 0} \log(f(t)) = 0 \implies \lim_{t \to 0} f(t) = 1 \text{ (Why?)}$$