💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here! # Find the limit. Use l'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. If l'Hospital's Rule doesn't apply, explain why.$\displaystyle \lim_{x\to 0} (1 - 2x)^{1/x}$

## $$\begin{array}{l}y=(1-2 x)^{1 / x} \Rightarrow \ln y=\frac{1}{x} \ln (1-2 x), \text { so } \lim _{x \rightarrow 0} \ln y=\lim _{x \rightarrow 0} \frac{\ln (1-2 x)}{x}=\lim _{x \rightarrow 0} \frac{-2 /(1-2 x)}{1}=-2 \Rightarrow \\\lim _{x \rightarrow 0}(1-2 x)^{1 / x}=\lim _{x \rightarrow 0} e^{\ln y}=e^{-2}\end{array}$$

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##### Top Calculus 2 / BC Educators ##### Catherine R.

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### Video Transcript

mhm. For this problem, we want to find the limit as X goes to zero. So um we start by taking the natural log of both sides. And in doing that we can use certain properties to get into a form that we want and that's going to be the natural log of 1 -2. X over X. However, when we take the limit we end up getting 0/0, which is the indeterminate form. So we're going to take the derivative of both the top and the bottom giving us um as a result will get the limit as X coke zero of one over 1 -2. X. Now, when we take the limit we end up getting negative two Because it's negative two times the limit. So we end up getting just negative too. Remember that that is going to be The limit as X approaches zero of the natural log of wine. So as a result of that, we end up um evaluating this and using what battles rule and that will give us That E to the -2 will be our final answer. California Baptist University

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Derivatives

Differentiation

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##### Top Calculus 2 / BC Educators ##### Catherine R.

Missouri State University   ##### Kristen K.

University of Michigan - Ann Arbor

Lectures

Join Bootcamp