Evaluating $\lim_{x\to0}(1-x)^{1/x}$

Question

Knowing $$\lim_{x\to0}(1+x)^{1/x}=e$$ is it possible to evaluate or is it obvious that $$\lim_{x\to0}(1-x)^{1/x}=e^{-1}$$

Answer 1

You can see the second limit as:

$\displaystyle\lim_{x\to 0} \frac{1}{(1-x)^{-1/x}}$

Then you use your first limit to get the asked for equality.

Answer 2

Guide:

• Do a substitution $y=-x$ and $\frac1x$ is a continuous function.

Answer 3

Not quite. Use L'hopital's rule. We see it approaches the indeterminate form $1^\infty$. So, taking the natural log, we see this limit is equal to $e^{\lim_{x\to 0}\log{(1-x)^{1/x}}}$. We see $\lim_{x\to 0}\log(1-x)^{1/x}=\lim_{x\to 0}\frac{\log(1-x)}{x}$. Once again, this approaches the indeterminate form $\frac{0}{0}$ so we take the derivative of the top and bottom to get $\lim_{x\to 0}\frac{-1}{1-x}=-1$. Thus, the solution is $e^{-1}$. Alternatively, you could make a subsitution $y=-x$ and use the already known limit to obtain the same result.