Limit of utility function in Ramsey-Cass-Koopmans model

I have given the utility function for the Ramsey-Cass-Koopmans model, as follows:

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

The claim is that as $$\theta \rightarrow 1$$, then the funciton is logarithmic, that is

$$\lim_{\theta\to 1} u(c) = ln(c)$$

I was told to use L'hôpital's rule, but I can't figure it out.

How can I prove this?

Thank you

The derivative of $$a^{x}$$ with respect to $$x$$ is equal to $$a^x \ln(a)$$.
As such, using the chain rule, the derivative of $$c^{1-\theta}$$ with respect to $$\theta$$ equals $$c^{1-\theta} \ln(c) (-1)$$
$$\lim_{\theta \to 1} \frac{c^{1-\theta}-1}{1 - \theta}\\ = \lim_{\theta \to 1} \frac{(c^{1-\theta}-1)'}{(1-\theta)'},\\ = \lim_{\theta \to 1} \frac{c^{1-\theta} \ln(c) (-1)}{(-1)},\\ = \frac{c^0 \ln(c)}{1} = \ln(c).$$
The answer of @tdm shows you how it is done using De L'Hospital's rule. If you want to avoid this rule you can write $$u(c)=\frac{c^{1-\theta}-1}{1-\theta}$$ and note that for $$\theta\ne 1$$ this implies $$u(1)=0$$. Differentiating with respect to $$c$$ then gives you $$u'(c)=c^{-\theta}$$ and therefore $$\lim_{\theta \to 1}u'(c)=c^{-1}$$. Then integrate again to get $$\lim_{\theta \to 1}u(c)=\ln(c)+K$$ and comparing for $$c=1$$ gives you $$K=0$$, so $$\lim_{\theta \to 1}u(c)=\ln(c)$$.