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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.

LuminousNutria
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  • Take $\log$ both side – Someone Apr 08 '19 at 19:28
  • https://math.stackexchange.com/questions/2493771/lhospitals-rule-for-lim-n-rightarrow-infty-an-bn-frac1n?rq=1 – Xander Henderson Apr 08 '19 at 19:31
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    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. – Xander Henderson Apr 08 '19 at 20:48

1 Answers1

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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 $\mathrm e^{\ell}$.

Bernard
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  • 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? – LuminousNutria Apr 08 '19 at 19:43
  • $-2 $ is the limit of the log. Hence the limit of the expression is … – Bernard Apr 08 '19 at 19:47
  • I wouldn't be able to figure that out from your hints Bernard. At least tell me a term I can look up. – LuminousNutria Apr 08 '19 at 19:49
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    $(1-2x)^{1/x}=\mathrm e^{\frac 1x\ln(1-2x)}$ since the exponential fiunction is the inverse function of the natural logarithm.. – Bernard Apr 08 '19 at 19:54
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    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 $-2$. – LuminousNutria Apr 08 '19 at 20:00
  • That's right. Note that for the limit of the log, you don't really need L'Hospital's rule, as the fraction is just a rate of variation, so the limit is just the derivative of the log at $x=0$. – Bernard Apr 08 '19 at 20:05