2
$\begingroup$

I am trying to find $\lim_{x \rightarrow0} (1-3 \cdot x)^{\frac{1}{x}}$

I thought about finding the limit of

$(1-3 \cdot x)^{\frac{1}{x}}= e^{\frac{\ln(1-3 \cdot x)}{x}}$

But that only works if $e^{\frac{\ln(1-3 \cdot x)}{x}}$ is continuous at $x=0$, which as far as I understand it is not. Am I right about that? And if I am, how do I find the limit?

Thanks!

  • 0
    Continuity **at** $0$ is irrelevant. To use many arguments, we do want the function to be continuous in a *deleted neighbourhood* of $0$. So there should exist an $\epsilon$ such that the function is continuous at all $x$ such that 0<|x|<\epsilon.2012-05-21

4 Answers 4

3

Instead of working with $e^{\frac1x\ln(1-3x)}$, work with $\frac{\ln(1-3x)}x$ directly: the hypotheses of l’Hospital’s rule are satisfied, so you can use it to find $\lim_{x\to 0}\frac{\ln(1-3x)}x=\lim_{x\to 0}\ln (1-3x)^{1/x}=\ln\lim_{x\to 0}(1-3x)^{1/x}\;,$ and then just exponentiate to get the desired limit.

2

Hint 1: What happens if you substitute $x=\frac 1n$? You will get a well known series.

Hint 2: One of the definitions of $e^a$ is given by $e^a=\lim_{n\to\infty}(1+\frac an)^n.$

  • 0
    I added a second hint. Does that help?2012-05-21
2

$\lim_{x \rightarrow0} (1-3 \cdot x)^{\frac{1}{x}}$ = $\lim_{x \rightarrow0} ((1+(-3) \cdot x)^{\frac{1}{-3x}})^{{-3}}$ = $e^{{-3}}$

  • 0
    Wait but that is only true if $e^{\frac{\ln(1-3x)}{x}}$ is continuous at $x=0$ isn't it?2012-05-21
2

Following the ideas of Simon and Prasad: if you know that $\,\displaystyle{\lim_{x\to\infty}\left(1+\frac{a}{f(x)}\right)^{f(x)}=e^a}\,$ for any function $\,f(x)\,$ s.t. $\,\displaystyle{\lim_{x\to\infty} f(x)=\infty}\,$ , then you're done as $\lim_{x\to 0}(1-3x)^{1/x}=\lim_{x\to 0}\left(1-\frac{3}{1/x}\right)^{1/x}=e^{-3}$