# L'Hospital's rule with indeterminate powers

I have no idea what to do here, I believe what I am doing is mathematically correct but the book does not give the answer I am getting.

$$\lim_{x\to 0} \; (1-2x)^{1/x}$$

For this I know that if I can make it the natural log (not sure why not just the log, but Stewart feels that isn't important to not so I will just blindly use the ln).

$$\frac{\ln(1-2x)}x$$

This is then very easy to calculate from here but I do not get the answer that I am suppose to. I get $2$ but I am suppose to get an answer of $e ^{-2}$

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1. What is the derivative of $\ln(1-2x)$? 2. Your limit calculation is for $\ln{f(x)}$, not your original function $f(x)$, what you need to do to $\ln{f(x)}$ to "get" $f(x)$? –  Salech Alhasov Apr 1 '12 at 2:28
I'll bet you did the derivative of $\ln(1-2x)$ incorrectly; did you remember to use the Chain Rule correctly? –  Arturo Magidin Apr 1 '12 at 2:28
Taking the derivative of a natural log is a lot cleaner. When you take the natural log of the limit, you are changing the value. Since $\ln x$ and $e^x$ are inverse functions, $e^{\ln x}=x$ Substitute your limit for $x$, then you can reverse the order of the limit and the natural log as Arturo did in his answer. Then you can use the necessary property of logarithms ($\ln a^b=b\ln a$)to get it into a form where you can apply l'Hopital's rule. –  Mike Apr 1 '12 at 6:34
I have a similar question, but with 2/x instead of 1/x. What do I do differently? –  cuabanana Apr 18 '13 at 22:44

For $x$ close to $0$, $1-2x$ is positive. So $$(1-2x)^{1/x} = e^{\ln(1-2x)/x}.$$ Since the exponential function is continuous, $$\lim_{x\to 0} e^{\ln(1-2x)/x} = e^{\scriptstyle\left(\lim\limits_{x\to 0}\ln(1-2x)/x\right)}$$ provided the latter limit exists. So this lets you change the original problem into the problem of determining whether $$\lim_{x\to 0}\frac{\ln(1-2x)}{x}$$ exists, and if so what the limit is.

(Alternatively, since $\ln$ is continuous, $$\lim_{x\to 0}\ln\left((1-2x)^{1/x}\right) = \ln\left(\lim_{x\to 0}(1-2x)^{1/x}\right)$$ so you can do the limit of the natural log instead).

Now, the limit $$\lim_{x\to 0}\frac{\ln(1-2x)}{x}$$ is an indeterminate of type $\frac{0}{0}$, so you can try using L'Hopital's rule. We get \begin{align*} \lim_{x\to 0}\frac{\ln(1-2x)}{x} &= \lim_{x\to 0}\frac{(\ln(1-2x))'}{x'} &\text{(L'Hopital's Rule)}\\ &= \lim_{x\to 0}\frac{\quad\frac{1}{1-2x}(1-2x)'\quad}{1}\\ &= \lim_{x\to 0}\frac{(1-2x)'}{1-2x} \\ &= \lim_{x\to 0}\frac{-2}{1-2x}\\ &= -2. \end{align*} Hence \begin{align*} \lim_{x\to 0}(1-2x)^{1/x} &= \lim_{x\to 0} e^{\ln(1-2x)/x}\\ &= e^{\lim\limits_{x\to 0}\ln(1-2x)/x}\\ &= e^{-2}. \end{align*}

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I don't understand where e is coming from. –  user138246 Apr 1 '12 at 3:30
@Jordan: By definition, for $a\gt 0$, we have $$a^b = e^{b\ln a}.$$ Alternatively, simply note that $e^{\ln x} = x$ for all $x\gt 0$, so $$a^b = e^{\ln(a^b)} = e^{b\ln(a)}.$$ –  Arturo Magidin Apr 1 '12 at 3:31

Let’s write out what’s really going on. You have $$L=\lim_{x\to 0}\;(1-2x)^{1/x}\;.$$ You take logs and use the continuity of the log function to get

$$\ln L=\ln\lim_{x\to 0}\;(1-2x)^{1/x}=\lim_{x\to 0}\;\frac1x\ln(1-2x)=\lim_{x\to 0}\;\frac{\ln(1-2x)}x\;.$$

Now you use l’Hospital’s rule to say that

$$\ln L=\lim_{x\to 0}\;\frac{\frac{-2}{1-2x}}1=\lim_{x\to 0}\frac{-2}{1-2x}=-2\;.$$

But this isn’t the original limit $L$: this is $\ln L$. To get $L$, you must exponentiate:

$$L=e^{\ln L}=e^{-2}\;.$$

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I don't know what you mean by L and how you are making the e ln L be -2. –  user138246 Apr 1 '12 at 3:29
@Jordan: $L$ is defined in the first displayed line: it’s simply a short name for the limit $\lim_{x\to 0}\;(1-2x)^{1/x}$. It’s easier to write $L$ than to keep writing $\lim_{x\to 0}\;(1-2x)^{1/x}$. I did not say that $e^{\ln L}$ was $-2$; $\ln L$ is $-2$, and $L$ is $e^{\ln L}$, so $L=e^{-2}$. –  Brian M. Scott Apr 1 '12 at 3:33