Evaluate $\lim_{x\rightarrow1}{\frac{1+\log{x}-e^{x-1}}{(x-1)^2}}$ using use L'Hospital's rule I want to evaluate the following limit
$$\lim_{x\rightarrow1}{\frac{1+\log{x}-e^{x-1}}{(x-1)^2}}$$
I use L'Hospital's rule:
$$\lim_{x\rightarrow1}{\frac{\frac{1}{x}-e^{x-1}}{2(x-1)}}$$
Now, here my textbook applies the L'Hospital's rule again. What I instead did was proceeding this way:
$$\lim_{x\rightarrow1}{\frac{\frac{1-xe^{x-1}}{x}}{2(x-1)}}=\lim_{x\rightarrow1}{\frac{1-xe^{x-1}}{2x(x-1)}}=\lim_{x\rightarrow1}{h(x)}$$
Now, 
$$h(x)\sim \frac{1-e^{x-1}}{2(x-1)}=\frac{-(e^{x-1}-1)}{2(x-1)}\sim\frac{-(x-1)}{2(x-1)}=-\frac 1 2$$
I think I did something wrong with the asymptotics because the solution is actually $-1$. Any hints?

You replaced $-xe^{x-1}$ by $-e^{x-1}$. The difference, $(1-x)e^{x-1}$ has the property that $\frac{(1-x)e^{x-1}}{2(x-1)}=-\frac{e^{x-1}}{2}\to -\frac12$ as $x\to 1$, which makes up for the shortfall in your solution.
