# Determining the limit as the denominator goes to zero

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

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

• What you wrote diverges to +/- infinity – lcv Oct 31 '19 at 2:27
• your equation is wrong, the nominator goes to 0 the denominator to 1 so the expression goes to infty – trula Oct 31 '19 at 2:29
• 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? – financial_physician Oct 31 '19 at 3:33

$$u(c_t) = \frac{c_t^{1-\gamma}-1}{1-\gamma}$$
Using L’Hopital’s rule for this $$0/0$$ indeterminate form, we have
$$\lim_{\gamma \to 1} u(c_t) = \lim_{\gamma \to 1}\frac{-\ln(c_t)e^{\ln(c_t)(1-\gamma)}}{-1} = \ln(c_t)$$