I am trying to find $\lim_{R \to 1} \frac{R^R x^{1-R}-1}{1-R}$, and the question explicitly states to use L'Hopital to find it. My current attempt is:

$\lim_{R \to 1} \frac{R^R x^{1-R}-1}{1-R} = \lim_{R \to 1} \frac{R^R(\ln R+1) x^{-R}(1-R)}{-1} = \lim_{R \to 1} \frac{R^R(\ln R+1) (R-1)}{x^R}$ (by L'Hopital)

I feel like I should now take logs, which yields:

$\lim_{R \to 1} \ln\left(\frac{R^R(\ln R+1) (R-1)}{x^R}\right) = \lim_{R \to 1} R \ln (R) + \ln(\ln R + 1) + \ln (R-1) - R \ln x = -\infty$.

So: $\lim_{R \to 1} \exp(-\infty) = 0$.

Is this correct? If not, where am I going wrong? (Hints appreciated maybe even more than answers!)

Thanks a million,

Best wishes,



Just $$ \lim_{R\to 1}\frac{R^R(\ln R+1)(R-1)}{x^R}=\lim_{R\to 1}\frac{1^1(\ln 1+1)(1-1)}{x^1}=0. $$ (your idea is correct but needless)


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