0
$\begingroup$

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 $e ^{-2}$

  • 0
    I have a similar question, but with 2/x instead of 1/x. What do I do differently?2013-04-18

3 Answers 3

5

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 $\begin{align*} \lim_{x\to 0}(1-2x)^{1/x} &= \lim_{x\to 0} e^{\ln(1-2x)/x}\\ &= e^{\lim\limits_{x\to 0}\ln(1-2x)/x}\\ &= e^{-2}. \end{align*}$

  • 0
    @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)}.$2012-04-01
4

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:

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

  • 0
    @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 $L=e^{-2}$.2012-04-01
0

Let us try another way since you already received good answers.

Consider $A= (1-2x)^{1/x}$ So, as you did, taking the logarithms $\ln(A)=\frac{\ln(1-2x)}x$ Now, remembering that, for small $y$, $\ln(1-y)\sim -y$, then $\ln(A)\sim\frac{-2x}x=-2$