Home » Blog » Compute the limit of the given function

Compute the limit of the given function

Evaluate the limit.

    \[ \lim_{x \to 0^-} \left( 1 - 2^x \right)^{\sin x}. \]

First, we use the definition of the exponential, and the continuity of e^x 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*}

Next, we rewrite the term inside the limit to get it into the form \infty / \infty 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*}

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 x =0 is equal to the value of the function. Since the numerator is 0 and the denominator is -1 when x = 0, we find that the limit is 0. Therefore,

    \begin{align*}   \lim_{x \to 0^-} \left( 1 - 2^x \right)^{\sin x} &= \exp \left( \log 2 \cdot 0 \right) \\  &= e^0 = 1. \end{align*}

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment):