This question already has an answer here:

I am trying to figure out how to make $\lim_{x \to 0} (1-2x)^{1/x}$ into $\frac{\infty}{\infty}$ or $\frac{0}{0}$ form for L'Hospital's Rule.

Using Desmos.com I have found that $(1-2x)^{1/x} \neq 1-2^{1/x}x^{1/x}$ and I'm not sure why that is either. That is the only thing I can think of to change the function's form.


marked as duplicate by Xander Henderson, LuminousNutria, Community Apr 8 at 20:55

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Take $\log$ both side $\endgroup$ – Mann Apr 8 at 19:28
  • $\begingroup$ math.stackexchange.com/questions/2493771/… $\endgroup$ – Xander Henderson Apr 8 at 19:31
  • 2
    $\begingroup$ It should be noted that the question @LuminousNutria marked as a dupe target is not an exact duplicate. However, I agree that it is, abstractly, a good duplicate target. The technique used there is identical: take a logarithm, work out the limit using L'H, then exponentiate. $\endgroup$ – Xander Henderson Apr 8 at 20:48

A suggestion:

Calculate the limit of the logarithm first: $$\lim_{x\to0}\ln\left(1-2x\right)^{1/x}=\lim_{x\to 0}\frac{\ln(1-2x)}x.$$

If you find the limit $\ell$ for the log, the limit of the expression is $\mathrm e^{\ell}$.

  • $\begingroup$ If I differentiate both sides I can get $\frac{-2}{1-2x}$ from this, which, when I replace $x$ with zero, becomes $-2$, but Symbolab.com says the answer is $\frac{1}{e^2}$. Is there something else I have to do? $\endgroup$ – LuminousNutria Apr 8 at 19:43
  • $\begingroup$ $-2 $ is the limit of the log. Hence the limit of the expression is … $\endgroup$ – Bernard Apr 8 at 19:47
  • $\begingroup$ I wouldn't be able to figure that out from your hints Bernard. At least tell me a term I can look up. $\endgroup$ – LuminousNutria Apr 8 at 19:49
  • 2
    $\begingroup$ $(1-2x)^{1/x}=\mathrm e^{\frac 1x\ln(1-2x)}$ since the exponential fiunction is the inverse function of the natural logarithm.. $\endgroup$ – Bernard Apr 8 at 19:54
  • 1
    $\begingroup$ Oh! I see! I get it now $\lim_{x \to 0} (1-2x)^{1/x} = e^{\lim_{x \to 0} \frac{\ln(1-2x)}{x}}$ The exponent of $e$ becomes $-2$. $\endgroup$ – LuminousNutria Apr 8 at 20:00

Not the answer you're looking for? Browse other questions tagged or ask your own question.