$\lim_{x\to0+}(\sinh(x))^{1/x}$
I started by setting $y=\frac{\sinh(x)}{x}$ and taking the natural logarithm of both sides and trying to solve the limit for $ln(y)$ but I got stuck trying to solve $\lim_{x\to0+}\frac{\ln(\sinh(x))}{x}$. Any ideas? Thanks in advanced.
$$\lim_{x\rightarrow 0^+}(\sinh(x))^{\frac{1}{x}}=\lim_{x\rightarrow 0^+}e^{\frac{1}{x}\ln(\sinh(x))}$$
$$\lim_{x\rightarrow 0^+}\frac{1}{x}=+\infty $$
$$\lim_{x\rightarrow 0^+}\ln(\sinh(x))=-\infty $$
so $$\lim_{x\rightarrow 0^+} e^{\frac{1}{x}\ln(\sinh(x))}=e^{-\infty}=0$$