How to use L'Hospital's rule when limit is undefined?

Question

I’m trying to find the following limit and I don’t know how I can use L’Hospital's rule since ln(0) is undefined. $$\lim_{t\rightarrow\tau} (\tau-t)\ln(\frac{\tau-t}{\tau})$$

Answer 1

Let $x = \tau - t$ so $x \rightarrow 0$, hence \begin{equation} \lim_{t\rightarrow\tau} (\tau-t)~ln(\frac{\tau-t}{\tau}) = \lim_{x\rightarrow 0} x \ln(\frac{x}{\tau}) = \lim_{x\rightarrow 0} x \ln x - x \ln \tau \end{equation} The first term goes to $0$ by L'Hopital and the second term is clearly going to zero.

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Why does $x\ln x \rightarrow 0$

\begin{equation} x\ln x = \frac{\ln x}{\frac{1}{x}} \end{equation} By L'Hopital we get

\begin{equation} x\ln x = \frac{\frac{1}{x}}{-\frac{1}{x^2}} = - x \rightarrow 0 \end{equation}

Answer 2

HINT

We have that

$$\lim_{t\rightarrow\tau^-} (\tau-t)\ln(\frac{\tau-t}{\tau})=\lim_{t\rightarrow\tau^-} \frac{\ln(\frac{\tau-t}{\tau})}{\frac1{\tau-t}}$$

Note that

• assuming $\tau>0$ the limit is well defined for $t\rightarrow\tau^-$
• by the change of variable $\tau-t=y \to 0^+$ the limit becomes

$$\lim_{t\rightarrow\tau^-} (\tau-t)\ln(\frac{\tau-t}{\tau})=\lim_{y\to 0^+} y\log\left(\frac y \tau\right)$$

which is a well known standard limit which can be evaluated without l'Hopital.

Indeed by $y=e^{-x}\to 0^+$ with $x\to \infty$ we have

$$y\log y=-\frac{x}{e^x} \to 0$$

indeed eventually $e^x>x^2$ and

$$\frac{x}{e^x}\le \frac 1 x \to 0$$