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It's been a while since I've taken an intro calculus class. Could someone remind me of this? I'm guessing it's La'Hopitals but a refresher would be super helpful

$ \lim_{\gamma\to1} \frac{c_{t}^{1-\gamma}}{1-\gamma} = ln(C_t) $

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    $\begingroup$ What you wrote diverges to +/- infinity $\endgroup$ – lcv Oct 31 '19 at 2:27
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    $\begingroup$ your equation is wrong, the nominator goes to 0 the denominator to 1 so the expression goes to infty $\endgroup$ – trula Oct 31 '19 at 2:29
  • $\begingroup$ hmmmm let me attach an image of what is written in the book I'm looking at in the book. Maybe I'm looking at it wrong? $\endgroup$ – financial_physician Oct 31 '19 at 3:33
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This is probably a misprint in the book. A power utility function should take the form

$$u(c_t) = \frac{c_t^{1-\gamma}-1}{1-\gamma}$$

Using L’Hopital’s rule for this $0/0$ indeterminate form, we have

$$\lim_{\gamma \to 1} u(c_t) = \lim_{\gamma \to 1}\frac{-\ln(c_t)e^{\ln(c_t)(1-\gamma)}}{-1} = \ln(c_t)$$

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  • $\begingroup$ Also known as a CRRA or isoelastic utility function. See Wikipedia reference under isoelastic utility. $\endgroup$ – RRL Oct 31 '19 at 3:56
  • $\begingroup$ you're the man. Thank you $\endgroup$ – financial_physician Oct 31 '19 at 3:59
  • $\begingroup$ You’re very welcome. $\endgroup$ – RRL Oct 31 '19 at 4:03

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