Calculate $\lim_{x \to 0^+} (1+\sin x)^{\frac{1}{\sqrt{x}}}$ I have this problem :
Calculate :
$$\lim_{x \to 0+} (1+\sin x)^{1/\sqrt{x}}$$
I don't really know to approach this.
Any ideas?
 A: The trick for handling limits that involve exponents is to take logs. Specifically:
$$
\begin{align*}
\text{Define }L&=\lim_{x\to 0^+}(1+\sin x)^{1/\sqrt{x}}\\
\log L &= \lim_{x\to 0^+}\frac{\log (1+\sin x)}{\sqrt{x}}\\
(y=\sqrt{x})\qquad&= \lim_{y\to 0^+}\frac{\log (1+\sin y^2)}{y}\\
\end{align*}
$$
Since $\log(1+z)=z+o(z)$ and $\sin(z)=z+o(z)$, it follows that
$$
\begin{align*}
\log L &= \lim_{y\to 0^+}\frac{y^2+o(y^2)}{y}\\
&=\lim_{y\to 0^+}y+o(y)\\
&=0.
\end{align*}
$$
Hence $L=1$.
EDIT: Had a mistake earlier.
A: Hint: Let $y = (1 + \sin(x))^{1/\sqrt{x}}$ so that $\ln(y) = \frac{\ln{(1+ \sin(x))}}{\sqrt{x}}$ by using the property $\ln(a^x) = x\ln(a)$ (which is true for all logarithmic functions, not just ln). Here, you can take the limit as $x \to 0^+$, and then exponente both sides to get the desired limit.
A: $\displaystyle f(x)=(1+\sin x)^\frac{1}{\sqrt{x}}=e^{\frac{1}{\sqrt{x}} \cdot\ln{(1+\sin x)}}$ but since $\ln(1+\sin x) \cdot\frac{1}{\sin x} \to 1 $ as $x \to 0^+$ then $f(x)=e^{\frac{\sin x}{\sqrt{x} \cdot \sin x} \cdot\ln{(1+\sin x)}} \to e^0=1$ because $\frac{\sin x}{\sqrt{x}}=\sqrt{x} \cdot\frac{\sin x}{x} \to 0$ as $x \to 0^+$ 
