Indeterminate Limit - L'Hospital's rule?

Question

I'm looking to evaluate algebraically the limit below which is an indeterminate form. I can get an answer from a graph, but I can't figure out how to algebraically evaluate the limit. I tried using logs to get it into the form where I can use L'Hospital's but that seems to make things worse?

\begin{equation} \lim_{x \to 1} (x-1)^{\sin(\pi x)} \end{equation}

The form I tried for L'Hospital's was: \begin{equation} e^{\lim_{x \to 1}{\frac{\ln(x-1)}{\csc(\pi x)}}} \end{equation} But the derivatives get messy really quickly. Any thoughts are appreciated.

Answer 1

Firstly note that the expression is well defined for $x>1$, then for $x \to 1^+$ we have

$$(x-1)^{\sin(\pi x)}=e^{\sin(\pi x) \log (x-1)} \to 1$$

indeed by $y=x-1 \to 0^+$

$$\sin(\pi x) \log (x-1)=\sin(\pi+\pi y)\log y=-\sin(\pi y)\log y=$$

$$=-\pi\frac{\sin(\pi y)}{\pi y}\cdot y\log y \to -\pi\cdot 1\cdot 0=0$$

since by l'Hopital (or by other methods) we have that

$$\lim_{y\to 0^+} y \log y =\lim_{y\to 0^+} \frac{ \log y}{\frac 1y}=\lim_{y\to 0^+} \frac{ \frac1y}{-\frac 1{y^2}}=\lim_{y\to 0^+} -y=0$$

Refer also to the related

• Evaluate $ \lim_{x \to 0} \left( {\frac{1}{x^2}} - {\frac{1} {\sin^2 x} }\right) $

Answer 2

Hint:

Compute the limit of the log in the first place For that, use substitution:

$$u=x-1\implies \ln(x-1)\sin\pi x=\ln u\sin(\pi u+\pi)=-\ln u\sin\pi u.$$ Can you proceed?

Answer 3

$$L=\lim_{x \to 1} (x-1)^{\sin(\pi x)}$$ $$\ln L=\lim_{x \to 1} {\sin(\pi x)} \ln (x-1)$$ Substitute $z=x-1$ $$\ln L=-\lim_{z \to 0} {\sin(\pi z)} \ln (z)$$ $$\ln L=-\lim_{z \to 0} \dfrac{\sin(\pi z)}{\pi z}\pi z \ln (z)$$ $$\ln L=-\lim_{z \to 0} \pi z \ln (z)$$ $$\ln L=-\pi\lim_{z \to 0} \dfrac {\ln (z)}{\dfrac 1z}$$ Apply l'Hôpital's rule: $$\ln L=\pi \lim_{z \to 0} z=0$$ $$\implies L=e^0=1$$