# Finding the limit of $(1-3 \cdot x)^{\frac{1}{x}}$

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!

• 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 $0<|x|<\epsilon$. – André Nicolas May 21 '12 at 14:02

## 4 Answers

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.

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

• Wait but that is only true if $e^{\frac{\ln(1-3x)}{x}}$ is continuous at $x=0$ isn't it? – yotamoo May 21 '12 at 10:06

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 $$e^a=\lim_{n\to\infty}(1+\frac an)^n.$$

• We haven't really dealt with series's, even though it seems like it converges to 1 as $x \rightarrow 0$ – yotamoo May 21 '12 at 9:55
• I added a second hint. Does that help? – Simon Markett May 21 '12 at 10:24

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 $$\lim_{x\to 0}(1-3x)^{1/x}=\lim_{x\to 0}\left(1-\frac{3}{1/x}\right)^{1/x}=e^{-3}$$