Evaluate the limit.
![\[ \lim_{x \to 0^-} \left( 1 - 2^x \right)^{\sin x}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-0045ebd98cfcea4c46589af011999c1c_l3.png "Rendered by QuickLaTeX.com")
First, we use the definition of the exponential, and the continuity of 
to write,
![\begin{align*} \lim_{x \to 0^-} \left( 1 - 2^x \right)^{\sin x} &= \lim_{x \to 0^-} e^{\sin x \log (1-2^x)} \\[9pt] &= \exp \left( \lim_{x \to 0^-} \left( \sin x \log (1-2^x) \right) \right). \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-0a94819bcf7da4633fa0fdc06e681622_l3.png "Rendered by QuickLaTeX.com")
Next, we rewrite the term inside the limit to get it into the form 
and apply L’Hopital’s rule,
![\begin{align*} \exp &\left( \lim_{x \to 0^-} \left( \sin x \log (1-2^x) \right) \right) \\[9pt] &= \exp \left( \lim_{x \to 0^-} \frac{\log (1-2^x)}{\frac{1}{\sin x}} \right) \\[9pt] &= \exp \left( \lim_{x \to 0^-} \frac{-\frac{2^x \log 2}{1-2^x}}{-\frac{\cos x}{\sin^2 x}} \right) &(\text{L'Hopital}) \\[9pt] &= \exp \left( \lim_{x \to 0^-} \frac{ \sin^2 x \cdot 2^x \log 2}{\cos x \cdot (1-2^x)} \right) \\[9pt] &= \exp \left( \log 2 \cdot \lim_{x \to 0^-} \left( \frac{2^x \sin x}{(1-2^x) \cos x} \right) \right) \\[9pt] &= \exp \left( \log 2 \cdot \lim_{x \to 0^-} \left( \frac{ 2^{x+1} \sin x \cos x + 2^x \log 2 \sin^2 x}{-2^x \cos x \log 2 - (1-2^x) \sin^2 x} \right) \right) &(\text{L'Hopital}). \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-0620401ac551d0ea8640a5a83dfb7127_l3.png "Rendered by QuickLaTeX.com")
But, this final limit is a quotient of continuous functions, and the denominator is not zero, so it is continuous. Therefore, the value of the limit at 
is equal to the value of the function. Since the numerator is 0 and the denominator is 
when 
, we find that the limit is 0. Therefore,
