What's the solution?

Calculus Level 1

$\large \lim_{x\to1 } (x-1)^{x-1} = \ ?$

1 -2 0 -1

2 solutions

Micah Wood
Nov 24, 2015

Let $t = x-1$ , so we have: $\lim_{x\to1}(x-1)^{x-1} = \lim_{t\to0}t^t$

Using the fact that $a^b = \exp( b\log a )$ , we have: $\lim_{t\to0}\exp( t \log t)$

Since $\exp$ is continuous function, we can factor it out limit: $\exp\left(\lim_{t\to0}t\log t\right) = \exp\left(\lim_{t\to0}\dfrac{\log t}{\frac1t}\right)$

Apply L'Hopital's Rule, we have: $\exp\left(\lim_{t\to0}-t\right) = \exp(0) = \boxed1$

The limit is indeed $1,$ but there is an issue with the limit as $t \rightarrow 0$ from the left side, since for $t \lt 0$ we end up with $\ln(t)$ having a complex component. L'Hopital's rule does apply for complex-valued functions under certain conditions, so care would have to be taken to be sure that these conditions are met in this case.

Brian Charlesworth - 5 years, 6 months ago

Without applying L'Hopital's Rule, so we have: $\frac { 1 }{ p } =t\Leftrightarrow \frac { 1 }{ t } =p,\quad t,p\in { \Re }^{ + }$

$exp\left( -{ \lim _{ p\rightarrow +\infty }{ \frac { \log { p } }{ p } } } \right) =exp\left( 0 \right) =1$

because the exponential grows faster than any power exponent.

Dinis Pereira - 5 years, 6 months ago

David Adut
Dec 3, 2015

What's the solution?

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